chemistry the central science 2

The units for reaction rate are usually molarity per second 1M>s2—that is, the change in concentration measured in molarity divided by a time interval measured in seconds.. In this cas

562 chapTer 13 properties of Solutions

Colloid stabilization has an interesting application in the human digestive system When fats in our diet reach the small intestine, a hormone causes the gallbladder to excrete a fluid called bile Among the components of bile are compounds that have chemical structures similar to sodium stearate; that is, they have a hydrophilic (polar) end and a hydrophobic (nonpolar) end These compounds emulsify the fats in the intes- tine and thus permit digestion and absorption of fat-soluble vitamins through the intes-

tinal wall The term emulsify means “to form an emulsion,” a suspension of one liquid in

another, with milk being one example (Table 13.5) A substance that aids in the tion of an emulsion is called an emulsifying agent If you read the labels on foods and other materials, you will find that a variety of chemicals are used as emulsifying agents These chemicals typically have a hydrophilic end and a hydrophobic end.

forma-Chemistry and Life

sickle-Cell Anemia

Our blood contains the complex protein hemoglobin, which carries

oxygen from the lungs to other parts of the body In the genetic

dis-ease sickle-cell anemia, hemoglobin molecules are abnormal and have

a lower solubility in water, especially in their unoxygenated form

Consequently, as much as 85% of the hemoglobin in red blood cells

crystallizes out of solution

The cause of the insolubility is a structural change in one part of

an amino acid Normal hemoglobin molecules contain an amino acid

that has a ¬ CH2CH2COOH group:

The polarity of the ¬ COOH group contributes to the solubility

of the hemoglobin molecule in water In the hemoglobin molecules of

sickle-cell anemia patients, the ¬ CH2CH2COOH chain is absent and

in its place is the nonpolar (hydrophobic) ¬ CH1CH322 group:

CH3CH

CH3

This change leads to the aggregation of the defective form of hemoglobin into particles too large to remain suspended in bio-logical fluids It also causes the cells to distort into the sickle shape shown in ▼ Figure 13.29 The sickled cells tend to clog capillar-

ies, causing severe pain, weakness, and the gradual deterioration

of vital organs The disease is hereditary, and if both parents carry the defective genes, it is likely that their children will possess only abnormal hemoglobin

You might wonder how it is that a life-threatening disease such

as sickle-cell anemia has persisted in humans through evolutionary time The answer in part is that people with the disease are far less susceptible to malaria Thus, in tropical climates rife with malaria, those with sickle-cell disease have lower incidence of this debilitating disease

▲ Figure 13.29 A scanning electron micrograph of normal (round) and sickle (crescent-shaped) red blood cells Normal red blood cells are

about 6 * 10-3 mm in diameter

Colloidal Motion in Liquids

We learned in Chapter 10 that gas molecules move at some average speed that depends inversely on their molar mass, in a straight line, until they collide with some-

thing The mean free path is the average distance molecules travel between collisions

(Section 10.8) Recall also that the kinetic-molecular theory of gases assumes that gas molecules are in continuous, random motion (Section 10.7)

SecTion 13.6 colloids 563

table 13.6 Calculated Mean Free Path, after one hour,

for uncharged Colloidal Spheres in Water at 20 °C

radius of sphere, nm Mean Free Path, mm

A 0.100-L solution is made by dissolving 0.441 g of CaCl21s2 in water (a) Calculate the osmotic

pressure of this solution at 27 °C, assuming that it is completely dissociated into its component

ions (b) The measured osmotic pressure of this solution is 2.56 atm at 27 °C Explain why it is

less than the value calculated in (a), and calculate the van’t Hoff factor, i, for the solute in this

solution (c) The enthalpy of solution for CaCl2 is ∆H = -81.3 kJ>mol If the final temperature

of the solution is 27 °C, what was its initial temperature? (Assume that the density of the

solu-tion is 1.00 g>mL, that its specific heat is 4.18 J>g@K, and that the solusolu-tion loses no heat to its

surroundings.)

SoLutIon

(a) The osmotic pressure is given by Equation 13.14, Π = iMRT We know the temperature,

T = 27 °C = 300 K, and the gas constant, R = 0.0821 L@atm/mol@K We can calculate the

molarity of the solution from the mass of CaCl2 and the volume of the solution:

Molarity = a0.441 g CaCl0.100 L 2b a1 mol CaCl110 g CaCl2

2b = 0.0397 mol CaCl2>LSoluble ionic compounds are strong electrolytes (Sections 4.1 and 4.3) Thus, CaCl2

consists of metal cations 1Ca2+2 and nonmetal anions 1Cl-2 When completely dissociated,

each CaCl2 unit forms three ions (one Ca2+ and two Cl-) Hence, the calculated osmotic

pressure is

Π = iMRT = 13210.0397 mol>L210.0821 L@atm>mol@K21300 K2 = 2.93 atm

(b) The actual values of colligative properties of electrolytes are less than those calculated

be-cause the electrostatic interactions between ions limit their independent movements In this

case, the van’t Hoff factor, which measures the extent to which electrolytes actually

dissoci-ate into ions, is given by

i = Π1calculated for nonelectrolyte2Π1measured2

10.0397 mol>L210.0821 L@atm>mol@K21300 K2 = 2.62Thus, the solution behaves as if the CaCl2 has dissociated into 2.62 particles instead of the

ideal 3

(c) If the solution is 0.0397 M in CaCl2 and has a total volume of 0.100 L, the number of moles

of solute is 10.100 L210.0397 mol>L2 = 0.00397 mol Hence, the quantity of heat

gener-ated in forming the solution is 10.00397 mol21-81.3 kJ>mol2 = -0.323 kJ The solution

S a M P L e

InteGratIVe exerCISe Putting Concepts together

Colloidal particles in a solution undergo random motion as a result of collisions

with solvent molecules Because the colloidal particles are massive in comparison

with solvent molecules, their movements from any one collision are very tiny

How-ever, there are many such collisions, and they cause a random motion of the entire

colloidal particle, called Brownian motion In 1905, Einstein developed an equation

for the average square of the displacement of a colloidal particle, a historically very

important development As you might expect, the larger the colloidal particle, the

shorter its mean free path in a given liquid (▼ table 13.6) Today, the understanding

of Brownian motion is applied to diverse problems in everything from cheese-making

to medical imaging.

564 chapTer 13 properties of Solutions

Chapter Summary and Key terms

tHe solution ProCess (seCtion 13.1) Solutions form when

one substance disperses uniformly throughout another The attractive

interaction of solvent molecules with solute is called solvation When

the solvent is water, the interaction is called hydration The dissolution

of ionic substances in water is promoted by hydration of the separated

ions by the polar water molecules The overall enthalpy change upon

solution formation may be either positive or negative Solution

forma-tion is favored both by a positive entropy change, corresponding to an

increased dispersal of the components of the solution, and by a

nega-tive enthalpy change, indicating an exothermic process

sAturAted solutions And soluBility (seCtion 13.2) The

equilibrium between a saturated solution and undissolved solute is

dynamic; the process of solution and the reverse process, crystallization,

occur simultaneously In a solution in equilibrium with undissolved

sol-ute, the two processes occur at equal rates, giving a saturated solution

If there is less solute present than is needed to saturate the solution, the

solution is unsaturated. When solute concentration is greater than the

equilibrium concentration value, the solution is supersaturated. This is an

unstable condition, and separation of some solute from the solution will

occur if the process is initiated with a solute seed crystal The amount of

solute needed to form a saturated solution at any particular temperature

is the solubility of that solute at that temperature

FACtors AFFeCtinG soluBility (seCtion 13.3) The solubility

of one substance in another depends on the tendency of systems to

become more random, by becoming more dispersed in space, and on

the relative intermolecular solute–solute and solvent–solvent energies

compared with solute–solvent interactions Polar and ionic solutes

tend to dissolve in polar solvents, and nonpolar solutes tend to

dis-solve in nonpolar dis-solvents (“like disdis-solves like”) Liquids that mix in

all proportions are miscible; those that do not dissolve significantly in

one another are immiscible Hydrogen-bonding interactions between

solute and solvent often play an important role in determining

solu-bility; for example, ethanol and water, whose molecules form

hydro-gen bonds with each other, are miscible The solubilities of gases in a

liquid are generally proportional to the pressure of the gas over the

solution, as expressed by Henry’s law: S g = kP g The solubilities of

most solid solutes in water increase as the temperature of the

solu-tion increases In contrast, the solubilities of gases in water generally

decrease with increasing temperature

exPressinG solution ConCentrAtions (seCtion 13.4)

Concentrations of solutions can be expressed quantitatively by eral different measures, including mass percentage [(mass solute/mass solution) * 100] parts per million (ppm), parts per billion (ppb), and

sev-mole fraction Molarity, M, is defined as sev-moles of solute per liter of

solution; molality, m, is defined as moles of solute per kilogram of

sol-vent Molarity can be converted to these other concentration units if the density of the solution is known

ColliGAtive ProPerties (seCtion 13.5) A physical property

of a solution that depends on the concentration of solute particles present, regardless of the nature of the solute, is a colligative property Colligative properties include vapor-pressure lowering, freezing-point lowering, boiling-point elevation, and osmotic pressure raoult’s law expresses the lowering of vapor pressure An ideal solution obeys Raoult’s law Differences in solvent–solute as compared with solvent–solvent and solute–solute intermolecular forces cause many solutions

to depart from ideal behavior

A solution containing a nonvolatile solute possesses a higher boiling point than the pure solvent The molal boiling-point-elevation constant,Kb, represents the increase in boiling point for a 1 m solu-

tion of solute particles as compared with the pure solvent Similarly, the molal freezing-point-depression constant, Kf, measures the lowering

of the freezing point of a solution for a 1 m solution of solute particles

The temperature changes are given by the equations ∆Tb = iKbm and

∆Tf= -iKf m where i is the van’t Hoff factor, which represents how many particles the solute breaks up into in the solvent When NaCl dissolves in water, two moles of solute particles are formed for each mole of dissolved salt The boiling point or freezing point is thus ele-vated or depressed, respectively, approximately twice as much as that

of a nonelectrolyte solution of the same concentration Similar erations apply to other strong electrolytes

consid-osmosis is the movement of solvent molecules through a able membrane from a less concentrated to a more concentrated solution This net movement of solvent generates an osmotic pressure, Π, which can be measured in units of gas pressure, such as atm The osmotic pres-

semiperme-sure of a solution is proportional to the solution molarity: Π = iMRT

Osmosis is a very important process in living systems, in which cell walls act as semipermeable membranes, permitting the passage of water but restricting the passage of ionic and macromolecular components

absorbs this heat, causing its temperature to increase The relationship between temperature change and heat is given by Equation 5.22:

q = 1specific heat21grams21∆T2 The heat absorbed by the solution is q = +0.323 kJ = 323 J The mass of the 0.100 L of

solution is 1100 mL211.00 g>mL2 = 100 g (to three significant figures) Thus, the ture change is

tem-Key equations 565

Colloids (seCtion 13.6) Particles that are large on the molecular

scale but still small enough to remain suspended indefinitely in a

sol-vent system form colloids, or colloidal dispersions Colloids, which are

intermediate between solutions and heterogeneous mixtures, have many

practical applications One useful physical property of colloids, the

scat-tering of visible light, is referred to as the tyndall effect Aqueous

col-loids are classified as hydrophilic or hydrophobic Hydrophilic colloids

are common in living organisms, in which large molecular aggregates (enzymes, antibodies) remain suspended because they have many polar,

or charged, atomic groups on their surfaces that interact with water Hydrophobic colloids, such as small droplets of oil, may remain in sus-pension through adsorption of charged particles on their surfaces.Colloids undergo Brownian motion in liquids, analogous to the random three-dimensional motion of gas molecules

Learning outcomes after studying this chapter, you should be able to:

• Describe how enthalpy and entropy changes affect solution

formation (Section 13.1)

• Describe the relationship between intermolecular forces and solubility,

including use of the “like dissolves like” rule (Sections 13.1 and 13.3)

• Describe the role of equilibrium in the solution process and its

relationship to the solubility of a solute (Section 13.2)

• Describe the effect of temperature on the solubility of solids and

gases in liquids (Section 13.3)

• Describe the relationship between the partial pressure of a gas and

its solubility (Section 13.3)

• Calculate the concentration of a solution in terms of molarity, molality,

mole fraction, percent composition, and parts per million and be able

to interconvert between them (Section 13.4)

• Describe what a colligative property is and explain the difference between the effects of nonelectrolytes and electrolytes on colliga- tive properties (Section 13.5)

• Calculate the vapor pressure of a solvent over a solution (Section 13.5)

• Calculate the boiling-point elevation and freezing-point depression

of a solution (Section 13.5)

• Calculate the osmotic pressure of a solution (Section 13.5)

• Explain the difference between a solution and a colloid (Section 13.6)

• Describe the similarities between the motions of gas molecules and the motions of colloids in a liquid (Section 13.6)

Key equations

• Sg = kPg [13.4] Henry’s law, which relates gas solubility to partial pressure

• Mass % of component = mass of component in soln

total mass of soln * 100 [13.5] Concentration in terms of mass percent

• ppm of component = mass of component in soln

total mass of soln * 106 [13.6] Concentration in terms of parts per million (ppm)

• Mole fraction of component = moles of component

total moles of all components [13.7] Concentration in terms of mole fraction

• Molarity = moles of solute

liters of soln [13.8] Concentration in terms of molarity

• Molality = moles of solute

kilograms of solvent [13.9] Concentration in terms of molality

• Psolution = XsolventP°solvent [13.10] Raoult’s law, calculating vapor pressure of solvent above a solution

• ∆Tb = iKbm [13.12] Calculating the boiling-point elevation of a solution

• ∆Tf= -iKfm [13.13] Calculating the freezing-point depression of a solution

• Π = ia V b n RT = iMRT [13.14] Calculating the osmotic pressure of a solution

566 chapTer 13 properties of Solutions

13.6 The solubility of Xe in water at 1 atm pressure and 20 °C is

approximately 5 * 10-3 M (a) Compare this with the

solu-bilities of Ar and Kr in water (Table 13.1) (b) What properties

of the rare gas atoms account for the variation in solubility? [Section 13.3]

13.7 The structures of vitamins E and B6 are shown below Predict which is more water soluble and which is more fat soluble Explain [Section 13.3]

13.8 You take a sample of water that is at room temperature and in

contact with air and put it under a vacuum Right away, you see bubbles leave the water, but after a little while, the bubbles stop As you keep applying the vacuum, more bubbles appear

A friend tells you that the first bubbles were water vapor, and the low pressure had reduced the boiling point of water, caus-ing the water to boil Another friend tells you that the first bubbles were gas molecules from the air (oxygen, nitrogen, and so forth) that were dissolved in the water Which friend is mostly likely to be correct? What, then, is responsible for the second batch of bubbles? [Section 13.4]

13.9 The figure shows two identical volumetric flasks containing the same solution at two temperatures

(a) Does the molarity of the solution change with the change

in temperature? Explain

(b) Does the molality of the solution change with the change

in temperature? Explain [Section 13.4]

Visualizing Concepts

13.1 Rank the contents of the following containers in order of

increasing entropy: [Section 13.1]

13.2 This figure shows the interaction of a cation with

surround-ing water molecules

+

(a) Which atom of water is associated with the cation? Explain.

(b) Which of the following explanations accounts for the

fact that the ion-solvent interaction is greater for Li+

than for K+?

a Li+ is of lower mass than K+

b The ionization energy of Li is higher than that for K.

c Li+ has a smaller ionic radius than K+

d Li has a lower density than K.

e Li reacts with water more slowly than K [Section 13.1]

13.3 Consider two ionic solids, both composed of singly-charged

ions, that have different lattice energies (a) Will the solids have

the same solubility in water? (b) If not, which solid will be more

soluble in water, the one with the larger lattice energy or the

one with the smaller lattice energy? Assume that solute-solvent

interactions are the same for both solids [Section 13.1]

13.4 Are gases always miscible with each other? Explain [Section 13.1]

13.5 Which of the following is the best representation of a

saturated solution? Explain your reasoning [Section 13.2]

exercises

exercises 567

(b) In making a solution, the enthalpy of mixing is always a

positive number

(c) An increase in entropy favors mixing.

13.14 Indicate whether each statement is true or false: (a) NaCl

dis-solves in water but not in benzene 1C6H62 because benzene is

denser than water (b) NaCl dissolves in water but not in

ben-zene because water has a large dipole moment and benben-zene

has zero dipole moment (c) NaCl dissolves in water but not

in benzene because the water–ion interactions are stronger than benzene–ion interactions

13.15 Indicate the type of solute–solvent interaction (Section 11.2) that should be most important in each of the following solu-

tions: (a) CCl4 in benzene 1C6H62, (b) methanol 1CH3OH2 in

water, (c) KBr in water, (d) HCl in acetonitrile 1CH3CN2

13.16 Indicate the principal type of solute–solvent interaction in

each of the following solutions and rank the solutions from

weakest to strongest solute–solvent interaction: (a) KCl in ter, (b) CH2Cl2 in benzene 1C6H62, (c) methanol 1CH3OH2 in water

13.17 An ionic compound has a very negative ∆Hsoln in water

(a) Would you expect it to be very soluble or nearly insoluble

in water? (b) Which term would you expect to be the largest

negative number: ∆Hsolvent, ∆Hsolute, or ∆Hmix?

13.18 When ammonium chloride dissolves in water, the solution

becomes colder (a) Is the solution process exothermic or dothermic? (b) Why does the solution form?

13.19 (a) In Equation 13.1, which of the enthalpy terms for

dissolv-ing an ionic solid would correspond to the lattice energy?

(b) Which energy term in this equation is always exothermic?

13.20 For the dissolution of LiCl in water, ∆Hsoln = -37 kJ>mol Which term would you expect to be the largest negative num-ber: ∆Hsolvent, ∆Hsolute, or ∆Hmix?

13.21 Two nonpolar organic liquids, hexane 1C6H142 and heptane 1C7H162, are mixed (a) Do you expect ∆Hsoln to be a large positive number, a large negative number, or close to zero?

Explain (b) Hexane and heptane are miscible with each other

in all proportions In making a solution of them, is the entropy

of the system increased, decreased, or close to zero, compared

to the separate pure liquids?

13.22 The enthalpy of solution of KBr in water is about

+198 kJ>mol Nevertheless, the solubility of KBr in water

is relatively high Why does the solution process occur even though it is endothermic?

Saturated Solutions; Factors affecting Solubility (Sections 13.2 and 13.3)

13.23 The solubility of Cr1NO323#9 H2O in water is 208 g per 100 g

of water at 15 °C A solution of Cr1NO323#9 H2O in water at

35 °C is formed by dissolving 324 g in 100 g of water When this solution is slowly cooled to 15 °C, no precipitate forms

(a) What term describes this solution? (b) What action might

you take to initiate crystallization? Use molecular-level cesses to explain how your suggested procedure works

13.24 The solubility of MnSO4#H2O in water at 20 °C is 70 g per

100 mL of water (a) Is a 1.22 M solution of MnSO4#H2O

in water at 20 °C saturated, supersaturated, or unsaturated?

(b) Given a solution of MnSO4#H2O of unknown tration, what experiment could you perform to determine whether the new solution is saturated, supersaturated, or unsaturated?

13.10 This portion of a phase diagram shows the vapor-pressure

curves of a volatile solvent and of a solution of that solvent

containing a nonvolatile solute (a) Which line represents the

solution? (b) What are the normal boiling points of the

sol-vent and the solution? [Section 13.5]

13.11 Suppose you had a balloon made of some highly flexible

semi-permeable membrane The balloon is filled completely with

a 0.2 M solution of some solute and is submerged in a 0.1 M

solution of the same solute:

0.1 M

0.2 M

Initially, the volume of solution in the balloon is 0.25 L

Assuming the volume outside the semipermeable membrane

is large, as the illustration shows, what would you expect for

the solution volume inside the balloon once the system has

come to equilibrium through osmosis? [Section 13.5]

13.12 The molecule n-octylglucoside, shown here, is widely used in

biochemical research as a nonionic detergent for “solubilizing”

large hydrophobic protein molecules What characteristics of

this molecule are important for its use in this way? [Section 13.6]

the Solution Process (Section 13.1)

13.13 Indicate whether each statement is true or false:

(a) A solute will dissolve in a solvent if solute–solute

interac-tions are stronger than solute-solvent interacinterac-tions

568 chapTer 13 properties of Solutions

13.25 By referring to Figure 13.15, determine whether the addition

of 40.0 g of each of the following ionic solids to 100 g of water

at 40 °C will lead to a saturated solution: (a) NaNO3, (b) KCl,

(c) K2Cr2O7, (d) Pb1NO322

13.26 By referring to Figure 13.15, determine the mass of each of the

following salts required to form a saturated solution in 250 g

of water at 30 °C: (a) KClO3, (b) Pb1NO322, (c) Ce21SO423

13.27 Consider water and glycerol, CH21OH2CH1OH2CH2OH

(a) Would you expect them to be miscible in all proportions?

Explain (b) List the intermolecular attractions that occur

be-tween a water molecule and a glycerol molecule

13.28 Oil and water are immiscible Which is the most likely reason?

(a) Oil molecules are denser than water (b) Oil molecules are

composed mostly of carbon and hydrogen (c) Oil molecules

have higher molar masses than water (d) Oil molecules have

higher vapor pressures than water (e) Oil molecules have

higher boiling points than water

13.29 Common laboratory solvents include acetone 1CH3COCH32,

methanol 1CH3OH2, toluene 1C6H5CH32, and water Which

of these is the best solvent for nonpolar solutes?

13.30 Would you expect alanine (an amino acid) to be more soluble

in water or in hexane? Explain

Alanine

13.31 (a) Would you expect stearic acid, CH31CH2216COOH, to be

more soluble in water or in carbon tetrachloride? Explain

(b) Which would you expect to be more soluble in water,

cyclohexane or dioxane? Explain

H2C

H2C

CH2

CH2O

13.32 Ibuprofen, widely used as a pain reliever, has a limited

solubil-ity in water, less than 1 mg>mL Which part of the molecule’s

structure (gray, white, red) contributes to its water solubility?

Which part of the molecule (gray, white, red) contributes to

its water insolubility?

Ibuprofen

13.33 Which of the following in each pair is likely to be more

soluble in hexane, C6H14: (a) CCl4 or CaCl2, (b) benzene

1C6H62 or glycerol, CH21OH2CH1OH2CH2OH, (c) octanoic

acid, CH3CH2CH2CH2CH2CH2CH2COOH, or acetic acid,

CH3COOH? Explain your answer in each case

13.34 Which of the following in each pair is likely to be more

solu-ble in water: (a) cyclohexane 1C6H122 or glucose 1C6H12O62,

(b) propionic acid 1CH3CH2COOH2 or sodium propionate 1CH3CH2COONa2, (c) HCl or ethyl chloride 1CH3CH2Cl2? Explain in each case

13.35 (a) Explain why carbonated beverages must be stored in

sealed containers (b) Once the beverage has been opened,

why does it maintain more carbonation when refrigerated than at room temperature?

13.36 Explain why pressure substantially affects the solubility of O2

in water but has little effect on the solubility of NaCl in water

13.37 The Henry’s law constant for helium gas in water at 30 °C

is 3.7 * 10-4 M>atm and the constant for N2 at 30 °C is 6.0 * 10-4 M>atm If the two gases are each present at 1.5

atm pressure, calculate the solubility of each gas

13.38 The partial pressure of O2 in air at sea level is 0.21 atm Using the data in Table 13.1, together with Henry’s law, calculate the molar concentration of O2 in the surface water of a mountain lake saturated with air at 20 °C and an atmospheric pressure

of 650 torr

Concentrations of Solutions (Section 13.4)

13.39 (a) Calculate the mass percentage of Na2SO4 in a solution containing 10.6 g of Na2SO4 in 483 g of water (b) An ore con-

tains 2.86 g of silver per ton of ore What is the concentration

of silver in ppm?

13.40 (a) What is the mass percentage of iodine in a solution

con-taining 0.035 mol I2 in 125 g of CCl4? (b) Seawater contains

0.0079 g of Sr2+ per kilogram of water What is the tion of Sr2+ in ppm?

13.41 A solution is made containing 14.6 g of CH3OH in 184 g of

H2O Calculate (a) the mole fraction of CH3OH, (b) the mass

percent of CH3OH, (c) the molality of CH3OH

13.42 A solution is made containing 20.8 g of phenol 1C6H5OH2 in

425 g of ethanol 1CH3CH2OH2 Calculate (a) the mole tion of phenol, (b) the mass percent of phenol, (c) the molal-

frac-ity of phenol

13.43 Calculate the molarity of the following aqueous solutions:

(a) 0.540 g of Mg1NO322 in 250.0 mL of solution, (b) 22.4 g of

LiClO4#3 H2O in 125 mL of solution, (c) 25.0 mL of 3.50 M

HNO3 diluted to 0.250 L

13.44 What is the molarity of each of the following solutions:

(a) 15.0 g of Al21SO423 in 0.250 mL solution, (b) 5.25 g of

Mn1NO322#2 H2O in 175 mL of solution, (c) 35.0 mL of

9.00 M H2SO4 diluted to 0.500 L?

13.45 Calculate the molality of each of the following solutions:

(a) 8.66 g of benzene 1C6H62 dissolved in 23.6 g of carbon rachloride 1CCl42, (b) 4.80 g of NaCl dissolved in 0.350 L of

tet-water

13.46 (a) What is the molality of a solution formed by dissolving

1.12 mol of KCl in 16.0 mol of water? (b) How many grams

of sulfur 1S82 must be dissolved in 100.0 g of naphthalene 1C10H82 to make a 0.12 m solution?

13.47 A sulfuric acid solution containing 571.6 g of H2SO4 per liter of solution has a density of 1.329 g>cm3 Calculate

such air in terms of (a) mol percentage, (b) molarity,

assum-ing 1 atm pressure and a body temperature of 37 °C?

Colligative Properties (Section 13.5)

13.61 You make a solution of a nonvolatile solute with a liquid vent Indicate whether each of the following statements is true

sol-or false (a) The freezing point of the solution is higher than that of the pure solvent (b) The freezing point of the solution

is lower than that of the pure solvent (c) The boiling point of the solution is higher than that of the pure solvent (d) The

boiling point of the solution is lower than that of the pure solvent

13.62 You make a solution of a nonvolatile solute with a liquid

sol-vent Indicate if each of the following statements is true or

false (a) The freezing point of the solution is unchanged by addition of the solvent (b) The solid that forms as the solu- tion freezes is nearly pure solute (c) The freezing point of the

solution is independent of the concentration of the solute

(d) The boiling point of the solution increases in proportion

to the concentration of the solute (e) At any temperature, the

vapor pressure of the solvent over the solution is lower than what it would be for the pure solvent

13.63 Consider two solutions, one formed by adding 10 g of glucose 1C6H12O62 to 1 L of water and the other formed by adding 10 g

of sucrose 1C12H22O112 to 1 L of water Calculate the vapor pressure for each solution at 20 °C; the vapor pressure of pure water at this temperature is 17.5 torr

13.64 (a) What is an ideal solution? (b) The vapor pressure of pure

water at 60 °C is 149 torr The vapor pressure of water over a solution at 60 °C containing equal numbers of moles of water and ethylene glycol (a nonvolatile solute) is 67 torr Is the so-lution ideal according to Raoult’s law? Explain

13.65 (a) Calculate the vapor pressure of water above a solution

prepared by adding 22.5 g of lactose 1C12H22O112 to 200.0 g

of water at 338 K (Vapor-pressure data for water are given

in Appendix B.) (b) Calculate the mass of propylene glycol

1C3H8O22 that must be added to 0.340 kg of water to reduce the vapor pressure by 2.88 torr at 40 °C

(a) the mass percentage, (b) the mole fraction, (c) the molality,

(d) the molarity of H2SO4 in this solution

13.48 Ascorbic acid 1vitamin C, C6H8O62 is a water-soluble

vita-min A solution containing 80.5 g of ascorbic acid dissolved in

210 g of water has a density of 1.22 g>mL at 55 °C Calculate

(a) the mass percentage, (b) the mole fraction, (c) the

molal-ity, (d) the molarity of ascorbic acid in this solution.

13.49 The density of acetonitrile 1CH3CN2 is 0.786 g>mL and the

density of methanol 1CH3OH2 is 0.791 g>mL A solution is

made by dissolving 22.5 mL of CH3OH in 98.7 mL of CH3CN

(a) What is the mole fraction of methanol in the solution?

(b) What is the molality of the solution? (c) Assuming that

the volumes are additive, what is the molarity of CH3OH in

the solution?

13.50 The density of toluene 1C7H82 is 0.867 g>mL, and the

den-sity of thiophene 1C4H4S2 is 1.065 g>mL A solution is made

by dissolving 8.10 g of thiophene in 250.0 mL of toluene

(a) Calculate the mole fraction of thiophene in the

solu-tion (b) Calculate the molality of thiophene in the solusolu-tion

(c) Assuming that the volumes of the solute and solvent are

additive, what is the molarity of thiophene in the solution?

13.51 Calculate the number of moles of solute present in each of

the following aqueous solutions: (a) 600 mL of 0.250 M SrBr2,

(b) 86.4 g of 0.180 m KCl, (c) 124.0 g of a solution that is 6.45%

glucose 1C6H12O62 by mass

13.52 Calculate the number of moles of solute present in each of the

fol-lowing solutions: (a) 255 mL of 1.50 M HNO31aq2, (b) 50.0 mg

of an aqueous solution that is 1.50 m NaCl, (c) 75.0 g of an

aqueous solution that is 1.50% sucrose 1C12H22O112 by mass

13.53 Describe how you would prepare each of the following

aqueous solutions, starting with solid KBr: (a) 0.75 L of

1.5 * 10-2 M KBr, (b) 125 g of 0.180 m KBr, (c) 1.85 L of a

solution that is 12.0% KBr by mass (the density of the

solu-tion is 1.10 g>mL), (d) a 0.150 M solusolu-tion of KBr that

con-tains just enough KBr to precipitate 16.0 g of AgBr from a

solution containing 0.480 mol of AgNO3

13.54 Describe how you would prepare each of the following

aque-ous solutions: (a) 1.50 L of 0.110 M 1NH422SO4 solution,

starting with solid 1NH422SO4; (b) 225 g of a solution that is

0.65 m in Na2CO3, starting with the solid solute; (c) 1.20 L of

a solution that is 15.0% Pb1NO322 by mass (the density of the

solution is 1.16 g>mL), starting with solid solute; (d) a 0.50 M

solution of HCl that would just neutralize 5.5 g of Ba1OH22

starting with 6.0 M HCl

13.55 Commercial aqueous nitric acid has a density of 1.42 g>mL

and is 16 M Calculate the percent HNO3 by mass in the

solution

13.56 Commercial concentrated aqueous ammonia is 28% NH3 by

mass and has a density of 0.90 g>mL What is the molarity of

this solution?

13.57 Brass is a substitutional alloy consisting of a solution of

cop-per and zinc A particular sample of red brass consisting of

80.0% Cu and 20.0% Zn by mass has a density of 8750 kg>m3

(a) What is the molality of Zn in the solid solution? (b) What

is the molarity of Zn in the solution?

13.58 Caffeine 1C8H10N4O22 is a stimulant found in coffee and tea

If a solution of caffeine in the solvent chloroform 1CHCl32

has a concentration of 0.0500 m, calculate (a) the percentage

of caffeine by mass, (b) the mole fraction of caffeine in the

solution

570 chapTer 13 properties of Solutions

point by 0.49 °C Calculate the approximate molar mass of adrenaline from this data

Adrenaline

13.80 Lauryl alcohol is obtained from coconut oil and is used to

make detergents A solution of 5.00 g of lauryl alcohol in 0.100 kg of benzene freezes at 4.1 °C What is the molar mass

of lauryl alcohol from this data?

13.81 Lysozyme is an enzyme that breaks bacterial cell walls A lution containing 0.150 g of this enzyme in 210 mL of solu-tion has an osmotic pressure of 0.953 torr at 25 °C What is the molar mass of lysozyme?

13.82 A dilute aqueous solution of an organic compound soluble

in water is formed by dissolving 2.35 g of the compound in water to form 0.250 L of solution The resulting solution has

an osmotic pressure of 0.605 atm at 25 °C Assuming that the organic compound is a nonelectrolyte, what is its molar mass?

[13.83] The osmotic pressure of a 0.010 M aqueous solution of CaCl2

is found to be 0.674 atm at 25 °C (a) Calculate the van’t Hoff

factor, i, for the solution (b) How would you expect the value

of i to change as the solution becomes more concentrated?

Explain

[13.84] Based on the data given in Table 13.4, which solution would

give the larger freezing-point lowering, a 0.030 m solution of NaCl or a 0.020 m solution of K2SO4? How do you explain the departure from ideal behavior and the differences observed between the two salts?

13.86 Choose the best answer: A colloidal dispersion of one

liq-uid in another is called (a) a gel, (b) an emulsion, (c) a foam, (d) an aerosol.

13.87 An “emulsifying agent” is a compound that helps stabilize a phobic colloid in a hydrophilic solvent (or a hydrophilic colloid in

hydro-a hydrophobic solvent) Which of the following choices is the best

emulsifying agent? (a) CH3COOH, (b) CH3CH2CH2COOH,

(c) CH31CH2211COOH, (d) CH31CH2211COONa

13.88 Aerosols are important components of the atmosphere

Does the presence of aerosols in the atmosphere increase or decrease the amount of sunlight that arrives at the Earth’s surface, compared to an “aerosol-free” atmosphere? Explain your reasoning

[13.89] Proteins can be precipitated out of aqueous solution by the addition of an electrolyte; this process is called “salting out”

13.66 (a) Calculate the vapor pressure of water above a solution

prepared by dissolving 28.5 g of glycerin 1C3H8O32 in 125 g

of water at 343 K (The vapor pressure of water is given in

Appendix B.) (b) Calculate the mass of ethylene glycol

1C2H6O22 that must be added to 1.00 kg of ethanol 1C2H5OH2

to reduce its vapor pressure by 10.0 torr at 35 °C The vapor

pressure of pure ethanol at 35 °C is 1.00 * 102 torr

[13.67] At 63.5 °C, the vapor pressure of H2O is 175 torr, and that of

ethanol 1C2H5OH2 is 400 torr A solution is made by mixing

equal masses of H2O and C2H5OH (a) What is the mole

frac-tion of ethanol in the solufrac-tion? (b) Assuming ideal-solufrac-tion

behavior, what is the vapor pressure of the solution at 63.5 °C?

(c) What is the mole fraction of ethanol in the vapor above the

solution?

[13.68] At 20 °C, the vapor pressure of benzene 1C6H62 is 75 torr,

and that of toluene 1C7H82 is 22 torr Assume that benzene

and toluene form an ideal solution (a) What is the

composi-tion in mole fraccomposi-tion of a solucomposi-tion that has a vapor pressure of

35 torr at 20 °C? (b) What is the mole fraction of benzene in

the vapor above the solution described in part (a)?

13.69 (a) Does a 0.10 m aqueous solution of NaCl have a higher

boiling point, a lower boiling point, or the same boiling point

as a 0.10 m aqueous solution of C6H12O6? (b) The

experi-mental boiling point of the NaCl solution is lower than that

calculated assuming that NaCl is completely dissociated in

solution Why is this the case?

13.70 Arrange the following aqueous solutions, each 10% by

mass in solute, in order of increasing boiling point: glucose

1C6H12O62, sucrose 1C12H22O112, sodium nitrate 1NaNO32

13.71 List the following aqueous solutions in order of

increas-ing boilincreas-ing point: 0.120 m glucose, 0.050 m LiBr, 0.050 m

Zn1NO322

13.72 List the following aqueous solutions in order of decreasing

freezing point: 0.040 m glycerin 1C3H8O32, 0.020 m KBr,

0.030 m phenol 1C6H5OH2

13.73 Using data from Table 13.3, calculate the freezing and boiling

points of each of the following solutions: (a) 0.22 m glycerol

1C3H8O32 in ethanol, (b) 0.240 mol of naphthalene 1C10H82

in 2.45 mol of chloroform, (c) 1.50 g NaCl in 0.250 kg of water,

(d) 2.04 g KBr and 4.82 g glucose 1C6H12O62 in 188 g of water

13.74 Using data from Table 13.3, calculate the freezing and boiling

points of each of the following solutions: (a) 0.25 m glucose in

ethanol; (b) 20.0 g of decane, C10H22, in 50.0 g CHCl3; (c) 3.50 g

NaOH in 175 g of water, (d) 0.45 mol ethylene glycol and

0.15 mol KBr in 150 g H2O

13.75 How many grams of ethylene glycol 1C2H6O22 must be

added to 1.00 kg of water to produce a solution that freezes at

13.78 Seawater contains 3.4 g of salts for every liter of solution

As-suming that the solute consists entirely of NaCl (in fact, over

90% of the salt is indeed NaCl), calculate the osmotic pressure

of seawater at 20 °C

13.79 Adrenaline is the hormone that triggers the release of extra

glucose molecules in times of stress or emergency A solution

of 0.64 g of adrenaline in 36.0 g of CCl4 elevates the boiling

additional exercises 571

same biochemistry class says that salting out works because the incoming ions adsorb tightly to the protein, making ion pairs on the protein surface, which end up giving the protein a zero net charge in water and therefore leading to precipitation Discuss these two hypotheses What kind of measurements would you need to make to distinguish between these two hypotheses?

13.90 Explain how (a) a soap such as sodium stearate stabilizes a

colloidal dispersion of oil droplets in water; (b) milk curdles

upon addition of an acid

the protein (a) Do you think that all proteins would be

pre-cipitated out to the same extent by the same concentration of

the same electrolyte? (b) If a protein has been salted out, are

the protein–protein interactions stronger or weaker than they

were before the electrolyte was added? (c) A friend of yours

who is taking a biochemistry class says that salting out works

because the waters of hydration that surround the protein

prefer to surround the electrolyte as the electrolyte is added;

therefore, the protein’s hydration shell is stripped away,

lead-ing to protein precipitation Another friend of yours in the

that seawater contains 13 ppt of gold, calculate the number of grams of gold contained in 1.0 * 103 gal of seawater

13.97 The maximum allowable concentration of lead in drinking

water is 9.0 ppb (a) Calculate the molarity of lead in a ppb solution (b) How many grams of lead are in a swimming

9.0-pool containing 9.0 ppb lead in 60 m3 of water?

13.98 Acetonitrile 1CH3CN2 is a polar organic solvent that dissolves

a wide range of solutes, including many salts The density of

a 1.80 M LiBr solution in acetonitrile is 0.826 g>cm3

Calcu-late the concentration of the solution in (a) molality, (b) mole fraction of LiBr, (c) mass percentage of CH3CN

13.99 A “canned heat” product used to warm buffet dishes consists

of a homogeneous mixture of ethanol 1C2H5OH2 and fin, which has an average formula of C24H50 What mass of

paraf-C2H5OH should be added to 620 kg of the paraffin to produce

8 torr of ethanol vapor pressure at 35 °C? The vapor pressure

of pure ethanol at 35 °C is 100 torr

13.100 A solution contains 0.115 mol H2O and an unknown number

of moles of sodium chloride The vapor pressure of the tion at 30 °C is 25.7 torr The vapor pressure of pure water at this temperature is 31.8 torr Calculate the number of grams

solu-of sodium chloride in the solution (Hint: Remember that

sodium chloride is a strong electrolyte.)

[13.101]Two beakers are placed in a sealed box at 25 °C One beaker

contains 30.0 mL of a 0.050 M aqueous solution of a

non-volatile nonelectrolyte The other beaker contains 30.0 mL of

a 0.035 M aqueous solution of NaCl The water vapor from

the two solutions reaches equilibrium (a) In which beaker

does the solution level rise, and in which one does it fall?

(b) What are the volumes in the two beakers when

equilib-rium is attained, assuming ideal behavior?

13.102 A car owner who knows no chemistry has to put antifreeze in

his car’s radiator The instructions recommend a mixture of 30% ethylene glycol and 70% water Thinking he will improve his protection he uses pure ethylene glycol, which is a liquid

at room temperature He is saddened to find that the tion does not provide as much protection as he hoped The pure ethylene glycol freezes solid in his radiator on a very cold day, while his neighbor, who did use the 30/70 mixture, has

solu-no problem Suggest an explanation

13.103 Calculate the freezing point of a 0.100 m aqueous solution of

K2SO4, (a) ignoring interionic attractions, and (b) taking

in-terionic attractions into consideration by using the van’t Hoff factor (Table 13.4)

13.104 Carbon disulfide 1CS22 boils at 46.30 °C and has a density of

1.261 g>mL (a) When 0.250 mol of a nondissociating solute

13.91 Butylated hydroxytoluene (BHT) has the following molecular

CH3

C

CH3

CH3

It is widely used as a preservative in a variety of foods,

in-cluding dried cereals Based on its structure, would you

ex-pect BHT to be more soluble in water or in hexane 1C6H142?

Explain

13.92 A saturated solution of sucrose 1C12H22O112 is made by

dis-solving excess table sugar in a flask of water There are 50 g

of undissolved sucrose crystals at the bottom of the flask in

contact with the saturated solution The flask is stoppered and

set aside A year later a single large crystal of mass 50 g is at

the bottom of the flask Explain how this experiment provides

evidence for a dynamic equilibrium between the saturated

so-lution and the undissolved solute

13.93 Most fish need at least 4 ppm dissolved O2 in water for

sur-vival (a) What is this concentration in mol>L? (b) What

par-tial pressure of O2 above water is needed to obtain 4 ppm O2

in water at 10 °C? (The Henry’s law constant for O2 at this

temperature is 1.71 * 10-3 mol>L@atm.)

13.94 The presence of the radioactive gas radon (Rn) in well

wa-ter presents a possible health hazard in parts of the United

States (a) Assuming that the solubility of radon in water

with 1 atm pressure of the gas over the water at 30 °C is

7.27 * 10-3 M, what is the Henry’s law constant for radon

in water at this temperature? (b) A sample consisting of

various gases contains 3.5 * 10-6 mole fraction of radon

This gas at a total pressure of 32 atm is shaken with water

at 30 °C Calculate the molar concentration of radon in

the water

13.95 Glucose makes up about 0.10% by mass of human blood

Calculate this concentration in (a) ppm, (b) molality

(c) What further information would you need to determine

the molarity of the solution?

13.96 The concentration of gold in seawater has been reported to

be between 5 ppt (parts per trillion) and 50 ppt Assuming

additional exercises

572 chapTer 13 properties of Solutions

0.036 g per 100 g of water at 25 °C The osmotic pressure of this solution is found to be 57.1 torr Assuming that molal-ity and molarity in such a dilute solution are the same and that the lithium salt is completely dissociated in the solu-

tion, determine an appropriate value of n in the formula for

the salt

is dissolved in 400.0 mL of CS2, the solution boils at 47.46 °C

What is the molal boiling-point-elevation constant for CS2?

(b) When 5.39 g of a nondissociating unknown is dissolved

in 50.0 mL of CS2, the solution boils at 47.08 °C What is the

molecular weight of the unknown?

[13.105]A lithium salt used in lubricating grease has the formula

LiCnH2n + 1O2 The salt is soluble in water to the extent of

(a) Account for the variations in heats of vaporization for

these substances, considering their relative intermolecular

forces (b) How would you expect the solubilities of these

sub-stances to vary in hexane as solvent? In ethanol? Use molecular forces, including hydrogen-bonding interactions where applicable, to explain your responses

[13.109] A textbook on chemical thermodynamics states, “The heat

of solution represents the difference between the lattice ergy of the crystalline solid and the solvation energy of the

en-gaseous ions.” (a) Draw a simple energy diagram to illustrate this statement (b) A salt such as NaBr is insoluble in most

polar nonaqueous solvents such as acetonitrile 1CH3CN2 or nitromethane 1CH3NO22, but salts of large cations, such as tetramethylammonium bromide 31CH324NBr4, are gener-ally more soluble Use the thermochemical cycle you drew

in part (a) and the factors that determine the lattice energy (Section 8.2) to explain this fact

13.110 (a) A sample of hydrogen gas is generated in a closed

con-tainer by reacting 2.050 g of zinc metal with 15.0 mL of

1.00 M sulfuric acid Write the balanced equation for the

reaction, and calculate the number of moles of hydrogen

formed, assuming that the reaction is complete (b) The

volume over the solution in the container is 122 mL late the partial pressure of the hydrogen gas in this volume

Calcu-at 25 °C, ignoring any solubility of the gas in the solution

(c) The Henry’s law constant for hydrogen in water at 25 °C

is 7.8 * 10-4 mol>L@atm Estimate the number of moles of hydrogen gas that remain dissolved in the solution What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hy-drogen in part (b)?

[13.111]The following table presents the solubilities of several gases in water at 25 °C under a total pressure of gas and

water vapor of 1 atm (a) What volume of CH41g2 under

standard conditions of temperature and pressure is

con-tained in 4.0 L of a saturated solution at 25 °C? (b)

Ex-plain the variation in solubility among the hydrocarbons listed (the first three compounds), based on their molecu-

lar structures and intermolecular forces (c) Compare the

solubilities of O2, N2, and NO, and account for the tions based on molecular structures and intermolecular

varia-forces (d) Account for the much larger values observed

for H2S and SO2 as compared with the other gases listed

(e) Find several pairs of substances with the same or nearly

the same molecular masses (for example, C2H4 and N2), and use intermolecular interactions to explain the differences in their solubilities

13.106 Fluorocarbons (compounds that contain both carbon and

fluo-rine) were, until recently, used as refrigerants The compounds

listed in the following table are all gases at 25 °C, and their

sol-ubilities in water at 25 °C and 1 atm fluorocarbon pressure are

given as mass percentages (a) For each fluorocarbon, calculate

the molality of a saturated solution (b) Explain why the molarity

of each of the solutions should be very close numerically to the

molality (c) Based on their molecular structures, account for the

differences in solubility of the four fluorocarbons (d) Calculate

the Henry’s law constant at 25 °C for CHClF2, and compare its

magnitude to that for N2 16.8 * 10-4 mol>L@atm2 Suggest a

reason for the difference in magnitude

Fluorocarbon Solubility (mass %)

[13.107]At ordinary body temperature 137 °C2, the solubility of N2 in

water at ordinary atmospheric pressure (1.0 atm) is 0.015 g>L

Air is approximately 78 mol % N2 (a) Calculate the number of

moles of N2 dissolved per liter of blood, assuming blood is a

sim-ple aqueous solution (b) At a depth of 100 ft in water, the external

pressure is 4.0 atm What is the solubility of N2 from air in blood

at this pressure? (c) If a scuba diver suddenly surfaces from this

depth, how many milliliters of N2 gas, in the form of tiny bubbles,

are released into the bloodstream from each liter of blood?

[13.108] Consider the following values for enthalpy of vaporization

1kJ>mol2 of several organic substances:

CH3C H 30.4O

Acetaldehyde

H2C CH2 28.5O

Ethylene oxide

CH3CCH3 32.0O

Acetone

CH2

24.7Cyclopropane

Integrative exercises

Design an experiment 573

bonds between acetone and chloroform molecules to explain

the deviation from ideal behavior (c) Based on the behavior

of the solution, predict whether the mixing of acetone and

chloroform is an exothermic 1∆Hsoln 6 02 or endothermic

1∆Hsoln 7 02 process (d) Would you expect the same

vapor-pressure behavior for acetone and chloromethane (CH3Cl)? Explain

13.114 Compounds like sodium stearate, called “surfactants” in

gen-eral, can form structures known as micelles in water, once the solution concentration reaches the value known as the critical micelle concentration (cmc) Micelles contain dozens to hun-dreds of molecules The cmc depends on the substance, the solvent, and the temperature

Surfactanttail Surfactanthead

cmc

At and above the cmc, the properties of the solution vary drastically

(a) The turbidity (the amount of light scattering) of

solu-tions increases dramatically at the cmc Suggest an

explana-tion (b) The ionic conductivity of the solution dramatically changes at the cmc Suggest an explanation (c) Chemists

have developed fluorescent dyes that glow brightly only when the dye molecules are in a hydrophobic environ-ment Predict how the intensity of such fluorescence would relate to the concentration of sodium stearate as the sodium stearate concentration approaches and then increases past the cmc

13.112 A small cube of lithium 1density = 0.535 g/cm32 measuring

1.0 mm on each edge is added to 0.500 L of water The

follow-ing reaction occurs:

2 Li1s2 + 2 H2O1l2 ¡ 2 LiOH1aq2 + H21g2

What is the freezing point of the resultant solution, assuming

that the reaction goes to completion?

[13.113] At 35 °C the vapor pressure of acetone, 1CH322CO, is 360

torr, and that of chloroform, CHCl3, is 300 torr Acetone and

chloroform can form very weak hydrogen bonds between one

another; the chlorines on the carbon give the carbon a

suf-ficient partial positive charge to enable this behavior:

CH3

CH3O

C

Cl Cl Cl

A solution composed of an equal number of moles of acetone

and chloroform has a vapor pressure of 250 torr at 35 °C

(a) What would be the vapor pressure of the solution if it

exhibited ideal behavior? (b) Use the existence of hydrogen

design an experiment

Based on Figure 13.18, you might think that the reason volatile solvent molecules in a solution

are less likely to escape to the gas phase, compared to the pure solvent, is because the solute

molecules are physically blocking the solvent molecules from leaving at the surface This is a

com-mon misconception Design an experiment to test the hypothesis that solute blocking of solvent

vaporization is not the reason that solutions have lower vapor pressures than pure solvents

14

Chemical Kinetics

the rusting of iron or the changing of color in leaves, occur relatively

slowly, requiring days, months, or years to complete Others, such as

the combustion reaction that generates the thrust for a rocket, as in the

chapter-opening photograph, happen much more rapidly As chemists,

we need to be concerned about the speed with which chemical

reactions occur as well as the products of the reactions For example,

14.3 ConCentration and rate Laws We show that the

effect of concentration on rate is expressed quantitatively by rate laws and show how rate laws and rate constants are determined

experimentally

14.4 the Change of ConCentration with time We learn that rate equations can be written to express how concentrations change with time and look at several classifications

of rate equations: zero-order, first-order, and second-order

reactions

14.1 faCtors that affeCt reaCtion rates We see that

four variables affect reaction rates: concentration, physical states

of reactants, temperature, and presence of catalysts These factors

can be understood in terms of the collisions among reactant

molecules that lead to reaction

14.2 reaCtion rates We examine how to express reaction

rates and how reactant disappearance rates and product

appearance rates are related to the reaction stoichiometry

What’s

ahead

▶ Launch of the Juno Spacecraft at

Cape Canaveral, Florida, in August 2011 The spacecraft is mounted to an Atlas V rocket, which at launch uses the very rapid combustion of kerosene and liquid oxygen

to generate its thrust

the chemical reactions that govern the metabolism of food, the transport of

essential nutrients, and your body’s ability to adjust to temperature changes (see

the Chemistry and Life box on the regulation of body temperature in Section 5.5)

all require that reactions occur with the appropriate speed Indeed, considerations

of the speeds of reactions are among the most important aspects of designing new

chemistry and chemical processes The area of chemistry concerned with the speeds, or

rates, of reactions is chemical kinetics.

So far, we have focused on the beginning and end of chemical reactions: We start

with certain reactants and see what products they yield This view is useful but does not

tell us what happens in the middle—that is, which chemical bonds are broken, which

are formed, and in what order these events occur The speed at which a chemical

reac-tion occurs is called the reacreac-tion rate Reacreac-tion rates can occur over very different time

14.7 CataLysis We end the chapter with a discussion of

how catalysts increase reaction rates, including a discussion of biological catalysts called enzymes.

14.5 temperature and rate We explore the effect of

temperature on rate In order to occur, most reactions require a

minimum input of energy called the activation energy.

14.6 reaCtion meChanisms We look more closely at

reaction mechanisms, the step-by-step molecular pathways

leading from reactants to products

576 chapTer 14 chemical Kinetics

scales (▲ Figure 14.1) To investigate how reactions happen, we must examine the tion rates and the factors that influence them Experimental information on the rate of

reac-a given rereac-action provides importreac-ant evidence threac-at helps us formulreac-ate reac-a rereac-action mechreac-a- nism, which is a step-by-step, molecular-level view of the pathway from reactants to

mecha-products.

Our goal in this chapter is to understand how to determine reaction rates and to consider the factors that control these rates What factors determine how rapidly food spoils, for instance? How does one design an automotive airbag that fills extremely rap- idly following a car crash? What determines the rate at which steel rusts? How can we remove hazardous pollutants in automobile exhaust before the exhaust leaves the tail- pipe? Although we will not address these specific questions, we will see that the rates of all chemical reactions are subject to the same principles.

Reaction Rates

Four factors affect the rate at which any particular reaction occurs:

1 Physical state of the reactants Reactants must come together to react The more

readily reactant molecules collide with one another, the more rapidly they react

Reactions may broadly classified as being either homogeneous, involving either all gases or all liquids, or as heterogeneous, in which reactants are in different phases

Under heterogeneous conditions, a reaction is limited by the area of contact of the reactants Thus, heterogeneous reactions that involve solids tend to proceed more rapidly if the surface area of the solid is increased For example, a medicine in the form of a fine powder dissolves in the stomach and enters the blood more quickly than the same medicine in the form of a tablet.

2 Reactant concentrations Most chemical reactions proceed more quickly if the

con-centration of one or more reactants is increased For example, steel wool burns only slowly in air, which contains 20% O2, but bursts into flame in pure oxygen (◀ Figure 14.2) As reactant concentration increases, the frequency with which the reactant molecules collide increases, leading to increased rates.

3 Reaction temperature Reaction rates generally increase as temperature is

in-creased The bacterial reactions that spoil milk, for instance, proceed more rapidly

at room temperature than at the lower temperature of a refrigerator Increasing temperature increases the kinetic energies of molecules (Section 10.7) As mol- ecules move more rapidly, they collide more frequently and with higher energy, leading to increased reaction rates.

15 s(30 million years)Time scale

▲ Figure 14.1 Reaction rates span an enormous range of time scales The absorption of light

by an atom or a molecule is complete within one femtosecond; explosions occur within seconds; corrosion can occur over years; and the weathering of rocks can occur over millions of years

Steel wool heated in air

(about 20% O2) glows red-hot

but oxidizes to Fe2O3 slowly

Red-hot steel wool in 100%

O2 burns vigorously, forming

Fe2O3 quickly

▲ Figure 14.2 Effect of concentration on

reaction rate The difference in behavior is

due to the different concentrations of O2 in

the two environments

Go FiGuRe

If a heated steel nail were placed in

pure O2, would you expect it to burn as

readily as the steel wool does?

secTIon 14.2 reaction rates 577

4 The presence of a catalyst Catalysts are agents that increase reaction rates without

themselves being used up They affect the kinds of collisions (and therefore alter

the mechanism) that lead to reaction Catalysts play many crucial roles in living

organisms, including ourselves.

On a molecular level, reaction rates depend on the frequency of collisions between

molecules The greater the frequency of collisions, the higher the reaction rate For a

col-lision to lead to a reaction, however, it must occur with sufficient energy to break bonds

and with suitable orientation for new bonds to form in the proper locations We will

consider these factors as we proceed through this chapter.

Give It Some Thought

In a reaction involving reactants in the gas state, how does increasing the partial

pressures of the gases affect the reaction rate?

The speed of an event is defined as the change that occurs in a given time interval, which

means that whenever we talk about speed, we necessarily bring in the notion of time

For example, the speed of a car is expressed as the change in the car’s position over a

certain time interval In the United States, the speed of cars is usually measured in units

of miles per hour—that is, the quantity that is changing (position measured in miles)

divided by a time interval (measured in hours).

Similarly, the speed of a chemical reaction—its reaction rate—is the change in the

concentration of reactants or products per unit of time The units for reaction rate are

usually molarity per second 1M>s2—that is, the change in concentration measured in

molarity divided by a time interval measured in seconds.

Let’s consider the hypothetical reaction A ¡ B, depicted in ▼ Figure 14.3

Each red sphere represents 0.01 mol of A, each blue sphere represents 0.01 mol of B,

and the container has a volume of 1.00 L At the beginning of the reaction, there is 1.00

mol A, so the concentration is 1.00 mol >L = 1.00 M After 20 s, the concentration of A

has fallen to 0.54 M and the concentration of B has risen to 0.46 M The sum of the

con-centrations is still 1.00 M because 1 mol of B is produced for each mole of A that reacts

After 40 s, the concentration of A is 0.30 M and that of B is 0.70 M.

▲ Figure 14.3 Progress of a hypothetical reaction a ¡ B The volume of the flask is 1.0 L.

578 chapTer 14 chemical Kinetics

The rate of this reaction can be expressed either as the rate of disappearance of

reactant A or as the rate of appearance of product B The average rate of appearance of

B over a particular time interval is given by the change in concentration of B divided by the change in time:

Average rate of appearance of B = change in concentration of B change in time

the 20-s interval from the beginning of the reaction 1t1 = 0 s to t2 = 20 s2 is

Average rate = 0.46 M - 0.00 M

20 s - 0 s = 2.3 * 10-2 M>s

We could equally well express the reaction rate in term of the reactant, A In this case, we would be describing the rate of disappearance of A, which we express as Average rate of disappearance of A = - change in concentration of A change in time

= - ∆3A4

Notice the minus sign in this equation, which we use to indicate that the

concen-tration of A decreases By convention, rates are always expressed as positive quantities

Because [A] decreases, ∆3A4 is a negative number The minus sign we put in the tion converts the negative ∆3A4 to a positive rate of disappearance.

equa-Because one molecule of A is consumed for every molecule of B that forms, the average rate of disappearance of A equals the average rate of appearance of B:

Average rate = - ∆3A4 ∆t = - 0.54 M - 1.00 M

20 s - 0 s = 2.3 * 10-2 M>s

From the data in Figure 14.3, calculate the average rate at which A disappears over the time interval from 20 s to 40 s

solution

analyze We are given the concentration of A at 20 s 10.54 M2 and at 40 s 10.30 M2 and asked to

calculate the average rate of reaction over this time interval

plan The average rate is given by the change in concentration, ∆3A4, divided by the change in time, ∆t

Because A is a reactant, a minus sign is used in the calculation to make the rate a positive quantity

(i) After 60 s there are 0.84 mol B in the flask

(ii) The decrease in the number of moles of A from t1 = 0 s to t2 = 20 s is greater than that from t1= 40 to t2= 60 s

(iii) The average rate for the reaction from t1 = 40 s to t2 = 60 s is 7.0 * 10-3 M>s

(a) Only one of the statements is true

(c) Statements (i) and (iii) are true

(e) All three statements are true.

(b) Statements (i) and (ii) are true

(d) Statements (ii) and (iii) are true

Practice exercise 2

Use the data in Figure 14.3 to calculate the average rate of appearance of B over the time interval from 0 s to 40 s

s a m P l e exeRCise 14.1 Calculating an average Rate of Reaction

secTIon 14.2 reaction rates 579

Change of Rate with Time

Now let’s consider the reaction between butyl chloride 1C4 H9Cl2 and water to form

butyl alcohol 1C4 H9OH2 and hydrochloric acid:

C4H9Cl1aq2 + H2O1l2 ¡ C4H9OH1aq2 + HCl1aq2 [14.3]

Suppose we prepare a 0.1000@M aqueous solution of C4H9Cl and then measure the

concentration of C4H9Cl at various times after time zero (which is the instant at which the

reactants are mixed, thereby initiating the reaction) We can use the

result-ing data, shown in the first two columns of ▲ table 14.1, to calculate the

average rate of disappearance of C4H9Cl over various time intervals; these

rates are given in the third column Notice that the average rate decreases

over each 50-s interval for the first several measurements and continues to

decrease over even larger intervals through the remaining measurements

It is typical for rates to decrease as a reaction proceeds because the

concen-tration of reactants decreases The change in rate as the reaction proceeds

is also seen in a graph of 3C4 H9Cl4 versus time (▶ Figure 14.4) Notice

how the steepness of the curve decreases with time, indicating a

decreas-ing reaction rate.

Instantaneous Rate

Graphs such as Figure 14.4 that show how the concentration of a

re-actant or product changes with time allow us to evaluate the

instanta-neous rate of a reaction, which is the rate at a particular instant during

the reaction The instantaneous rate is determined from the slope of the

curve at a particular point in time We have drawn two tangent lines

in Figure 14.4, a dashed line running through the point at t = 0 s

and a solid line running through the point at t = 600 s The slopes

of these tangent lines give the instantaneous rates at these two time

points.* To determine the instantaneous rate at 600 s, for example, we

construct horizontal and vertical lines to form the blue right triangle

in Figure 14.4 The slope of the tangent line is the ratio of the height of

the vertical side to the length of the horizontal side:

Instantaneous rate = - ∆3C4H9Cl4 ∆t = - 10.017 - 0.0422 M 1800 - 4002 s

= 6.3 * 10-5 M>s

table 14.1 Rate data for Reaction of C 4 h 9 Cl with Water

time, t 1s2 3C4 h 9 Cl4 1M2 average Rate 1M,s2

0.0100.0200.0300.0400.0500.0600.0700.0800.0900.100

time t.

*You may wish to review graphical determination of slopes in Appendix A If you are familiar with calculus, you

may recognize that the average rate approaches the instantaneous rate as the time interval approaches zero This

limit, in the notation of calculus, is the negative of the derivative of the curve at time t, -d3C4H9Cl4>dt.

580 chapTer 14 chemical Kinetics

In discussions that follow, the term rate means instantaneous rate unless

indi-cated otherwise The instantaneous rate at t = 0 is called the initial rate of the reaction

To understand the difference between average and instantaneous rates, imagine you have just driven 98 mi in 2.0 h Your average speed for the trip is 49 mi>hr, but your instantaneous speed at any moment during the trip is the speedometer reading at that moment.

Using Figure 14.4, calculate the instantaneous rate of disappearance of C4H9Cl at t = 0 s (the

initial rate)

solution

analyze We are asked to determine an instantaneous rate from a graph of reactant concentration versus time

plan To obtain the instantaneous rate at t = 0 s, we must determine the slope of the curve at

t = 0 The tangent is drawn on the graph as the hypotenuse of the tan triangle The slope of this

straight line equals the change in the vertical axis divided by the corresponding change in the horizontal axis (which, in the case of this example, is the change in molarity over change in time)

Solve The tangent line falls from 3C4H9Cl4 = 0.100 M to 0.060 M in the time change from 0 s

to 210 s Thus, the initial rate is

Give It Some Thought

In Figure 14.4, order the following three rates from fastest to slowest: (i) The average rate of the reaction between 0 s and 600 s, (ii) the instantaneous rate at

t = 0 s, and (iii) the instantaneous rate at t = 600 s You should not have to do

any calculations

Reaction Rates and Stoichiometry During our discussion of the hypothetical reaction A ¡ B, we saw that the stoi- chiometry requires that the rate of disappearance of A equal the rate of appearance

of B Likewise, the stoichiometry of Equation 14.3 indicates that 1 mol of C4H9OH

is produced for each mole of C4H9Cl consumed Therefore, the rate of appearance of C4H9OH equals the rate of disappearance of C4H9Cl:

Rate = - ∆3C4H9Cl4 ∆t = ∆3C4H9OH4

∆t What happens when the stoichiometric relationships are not one-to-one? For

example, consider the reaction 2 HI1g2 ¡ H21g2 + I21g2 We can measure either

the rate of disappearance of HI or the rate of appearance of either H2 or I2 Because 2 mol of HI disappears for each mole of H2 or I2 that forms, the rate of disappearance of

HI is twice the rate of appearance of either H2 or I2 How do we decide which number

to use for the rate of the reaction? Depending on whether we monitor HI, I2, or H2, the rates can differ by a factor of 2 To fix this problem, we need to take into account the reaction stoichiometry To arrive at a number for the reaction rate that does not

secTIon 14.3 concentration and rate Laws 581

depend on which component we measured, we must divide the rate of disappearance of

HI by 2 (its coefficient in the balanced chemical equation):

When we speak of the rate of a reaction without specifying a particular reactant or

product, we utilize the definition in Equation 14.4.*

*Equation 14.4 does not hold true if substances other than C and D are formed in significant amounts For

example, sometimes intermediate substances build in concentration before forming the final products In

that case, the relationship between the rate of disappearance of reactants and the rate of appearance of

prod-ucts is not given by Equation 14.4 All reactions whose rates we consider in this chapter obey Equation 14.4

One way of studying the effect of concentration on reaction rate is to determine the

way in which the initial rate of a reaction depends on the initial concentrations For

example, we might study the rate of the reaction

NH4+1aq2 + NO2 -1aq2 ¡ N21g2 + 2 H2O1l2

solution

analyze We are given a balanced chemical equation and asked to relate

the rate of appearance of the product to the rate of disappearance of the

reactant

plan We can use the coefficients in the chemical equation as shown in

Equation 14.4 to express the relative rates of reactions

Solve

(a) Using the coefficients in the balanced equation and the

relationship given by Equation 14.4, we have:

Rate = - 12 ∆3O∆t34 = 1

3

∆3O24

∆t

(b) Solving the equation from part (a) for the rate at which

O3 disappears, - ∆3O34>∆t, we have:

check We can apply a stoichiometric factor to convert the

O2 formation rate to the O3 disappearance rate:

- ∆3O∆t34 = a6.0 * 10-5 mol Os2>Lb a2 mol O3 mol O3

2b = 4.0 * 10-5 mol Os3>L = 4.0 * 10-5M>s

s a m P l e

exeRCise 14.3 Relating Rates at Which Products appear and Reactants disappear

(a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the

reaction 2 O31g2 ¡ 3 O21g2?

(b) If the rate at which O2 appears, ∆3O24>∆t, is 6.0 * 10-5 M>s at a particular instant, at what

rate is O3 disappearing at this same time, - ∆3O34>∆t?

Practice exercise 1

At a certain time in a reaction, substance A is disappearing at

a rate of 4.0 * 10-2 M>s, substance B is appearing at a rate

of 2.0 * 10-2 M>s, and substance C is appearing at a rate of

6.0 * 10-2 M>s Which of the following could be the

stoichiometry for the reaction being studied?

(a) 2A + B ¡ 3C (b) A ¡ 2B + 3C (c) 2A ¡ B + 3C (d) 4A ¡ 2B + 3C (e) A + 2B ¡ 3C

582 chapTer 14 chemical Kinetics

by measuring the concentration of NH4+ or NO2- as a function of time or by ing the volume of N2 collected as a function of time Because the stoichiometric coef- ficients on NH4+, NO2-, and N2 are the same, all of these rates are the same.

measur-▲ table 14.2 shows that changing the initial concentration of either reactant changes the initial reaction rate If we double 3NH4+4 while holding 3NO2-4 constant, the rate doubles (compare experiments 1 and 2) If we increase 3NH4+4 by a factor of 4 but leave 3NO2-4 unchanged (experiments 1 and 3), the rate changes by a factor of 4, and

so forth These results indicate that the initial reaction rate is proportional to 3NH4+4 When 3NO2-4 is similarly varied while 3NH4+4 is held constant, the rate is affected in the same manner Thus, the rate is also directly proportional to the concentration of 3NO2-4.

table 14.2 Rate data for the Reaction of ammonium and nitrite ions in Water at 25 °C

experiment number initial nh 4

passes, and c is the molar concentration of the absorbing substance

Thus, the concentration is directly proportional to absorbance Many chemical and pharmaceutical companies routinely use Beer’s law to calculate the concentration of purified solutions of the compounds that they make In the laboratory portion of your course, you may very well perform one or more experiments in which you use Beer’s law to relate absorption of light to concentration

Related Exercises: 14.101, 14.102, Design an Experiment

a Closer look

Using Spectroscopic Methods to

Measure Reaction Rates: Beer’s Law

A variety of techniques can be used to monitor reactant and product

concentration during a reaction, including spectroscopic methods,

which rely on the ability of substances to absorb (or emit) light

Spec-troscopic kinetic studies are often performed with the reaction

mix-ture in the sample compartment of a spectrometer, an instrument that

measures the amount of light transmitted or absorbed by a sample at

different wavelengths For kinetic studies, the spectrometer is set to

measure the light absorbed at a wavelength characteristic of one of the

reactants or products In the decomposition of HI(g) into H21g2 and

I21g2, for example, both HI and H2 are colorless, whereas I2 is violet

During the reaction, the violet color of the reaction mixture gets more

intense as I2 forms Thus, visible light of appropriate wavelength can

be used to monitor the reaction (▶ Figure 14.5)

▶ Figure 14.6 shows the components of a spectrometer The

spectrometer measures the amount of light absorbed by the sample by

comparing the intensity of the light emitted from the light source with

the intensity of the light transmitted through the sample, for various

wavelengths As the concentration of I2 increases and its color becomes

more intense, the amount of light absorbed by the reaction mixture

in-creases, as Figure 14.5 shows, causing less light to reach the detector

How can we relate the amount of light detected by the spectrometer

to the concentration of a species? A relationship called Beer’s law gives us

a direct route to the information we seek Beer’s law connects the amount

of light absorbed to the concentration of the absorbing substance:

A = ebc [14.5]

In this equation, A is the measured absorbance, e is the

extinc-tion coefficient (a characteristic of the substance being monitored at a

given wavelength of light), b is the path length through which the light

▲ Figure 14.5 Visible spectra of l 2 at different concentrations.

0.00.20.4

concentration increases

secTIon 14.3 concentration and rate Laws 583

We express the way in which the rate depends on the reactant concentrations by

the equation

An equation such as Equation 14.6, which shows how the rate depends on reactant

concentrations, is called a rate law For the general reaction

a A + b B ¡ c C + d D

the rate law generally has the form

Notice that only the concentrations of the reactants generally appear in the rate law The

constant k is called the rate constant The magnitude of k changes with temperature

and therefore determines how temperature affects rate, as we will see in Section 14.5

The exponents m and n are typically small whole numbers As we will learn shortly, if

we know m and n for a reaction, we can gain great insight into the individual steps that

occur during the reaction.

Give It Some Thought

How do reaction rate, rate law, and rate constant differ?

Once we know the rate law for a reaction and the reaction rate for a set of

reac-tant concentrations, we can calculate the value of k For example, using the values for

experiment 1 in Table 14.2, we can substitute into Equation 14.6:

5.4 * 10-7 M>s = k10.0100 M210.200 M2

k = 10.0100 M210.200 M2 5.4 * 10-7 M>s = 2.7 * 10-4 M-1s-1

You should verify that this same value of k is obtained using any of the other

experi-mental results in Table 14.2.

Once we have both the rate law and the k value for a reaction, we can

calcu-late the reaction rate for any set of concentrations For example, using Equation

14.7 with k = 2.7 * 10-4 M-1 s-1, m = 1, and n = 1, we can calculate the rate for

3NH4+4 = 0.100 M and 3NO2-4 = 0.100 M:

Rate = 12.7 * 10-4M-1 s-1210.100 M210.100 M2 = 2.7 * 10-6M>s

Give It Some Thought

Does the rate constant have the same units as the rate?

▲ Figure 14.6 Components of a spectrometer

Source Lenses/slits/

collimators Monochromator(selects wavelength) Sample Detector Computer

584 chapTer 14 chemical Kinetics

Reaction Orders: The Exponents in the Rate Law The rate law for most reactions has the form

The exponents m and n are called reaction orders For example, consider again the

rate law for the reaction of NH4+ with NO2-:

Rate = k3NH4+43NO2-4 Because the exponent of 3NH4+4 is 1, the rate is first order in NH4+ The rate is also first order in NO2- (The exponent 1 is not shown in rate laws.) The overall reaction order is the sum of the orders with respect to each reactant represented in the rate

law Thus, for the NH4+ - NO2- reaction, the rate law has an overall reaction order of

1 + 1 = 2, and the reaction is second order overall.

The exponents in a rate law indicate how the rate is affected by each reactant centration Because the rate at which NH4+ reacts with NO2- depends on 3NH4+4 raised

con-to the first power, the rate doubles when 3NH4+4 doubles, triples when 3NH4+4 triples, and so forth Doubling or tripling 3NO2-4 likewise doubles or triples the rate If a rate law is second order with respect to a reactant, 3A42, then doubling the concentration of that substance causes the reaction rate to quadruple because 3242 = 4, whereas tripling the concentration causes the rate to increase ninefold: 3342 = 9.

The following are some additional examples of experimentally determined rate laws:

CHCl31g2 + Cl21g2 ¡ CCl41g2 + HCl1g2 Rate = k3CHCl343Cl241>2 [14.11] Although the exponents in a rate law are sometimes the same as the coefficients in the balanced equation, this is not necessarily the case, as Equations 14.9 and 14.11 show For

any reaction, the rate law must be determined experimentally In most rate laws,

reac-tion orders are 0, 1, or 2 However, we also occasionally encounter rate laws in which the reaction order is fractional (as is the case with Equation 14.11) or even negative.

Give It Some Thought

The experimentally determined rate law for the reaction

2 NO1g2 + 2 H21g2 ¡ N21g2 + 2 H2O1g2 is rate = k3NO423H24

(a) What are the reaction orders in this rate law?

(b) Would the reaction rate increase more if we doubled the concentration of NO

or the concentration of H2?

s a m P l e

exeRCise 14.4 Relating a Rate law to the effect of Concentration on Rate

Consider a reaction A + B ¡ C for which rate = k3A43B42 Each of the following boxes

repre-sents a reaction mixture in which A is shown as red spheres and B as purple ones Rank these

mix-tures in order of increasing rate of reaction

secTIon 14.3 concentration and rate Laws 585

Magnitudes and Units of Rate Constants

If chemists want to compare reactions to evaluate which ones are relatively fast and

which ones are relatively slow, the quantity of interest is the rate constant A good

gen-eral rule is that a large value of k (∼109 or higher) means a fast reaction and a small

value of k (10 or lower) means a slow reaction.

Give It Some Thought

Suppose the reactions A ¡ B and X ¡ Y have the same value of k When

3A4 = 3X4, will the two reactions necessarily have the same rate?

The units of the rate constant depend on the overall reaction order of the rate law

In a reaction that is second order overall, for example, the units of the rate constant

must satisfy the equation:

Units of rate = 1units of rate constant21units of concentration22

Hence, in our usual units of molarity for concentration and seconds for time,

analyze We are given three boxes containing different numbers of

spheres representing mixtures containing different reactant

con-centrations We are asked to use the given rate law and the

com-positions of the boxes to rank the mixtures in order of increasing

reaction rates

plan Because all three boxes have the same volume, we can put the

number of spheres of each kind into the rate law and calculate the rate

for each box

Solve Box 1 contains 5 red spheres and 5 purple spheres, giving the

The slowest rate is 63k (Box 2), and the highest is 147k (Box 3) Thus,

the rates vary in the order 2 6 1 6 3

check Each box contains 10 spheres The rate law indicates that in this case [B] has a greater influence on rate than [A] because B has a larger reaction order Hence, the mixture with the highest concentra-tion of B (most purple spheres) should react fastest This analysis con-firms the order 2 6 1 6 3

Practice exercise 1

Suppose the rate law for the reaction in this Sample Exercise were

rate = k3A423B4 What would be the ordering of the rates for the three mixtures shown above, from slowest to fastest?

(a) 1 6 2 6 3 (b) 1 6 3 6 2 (c) 3 6 2 6 1 (d) 2 6 1 6 3 (e) 3 6 1 6 2

Practice exercise 2

Assuming that rate = k3A43B4, rank the mixtures represented in

this Sample Exercise in order of increasing rate

solution

analyze We are given two rate laws and asked to express (a) the overall

reaction order for each and (b) the units for the rate constant for the first

reaction

plan The overall reaction order is the sum of the exponents in the rate

law The units for the rate constant, k, are found by using the normal

units for rate 1M>s2 and concentration (M) in the rate law and

apply-ing algebra to solve for k.

Solve

(a) The rate of the reaction in Equation 14.9 is first order in N2O5

and first order overall The reaction in Equation 14.11 is first

s a m P l e

exeRCise 14.5 determining Reaction orders and units for Rate Constants

(a) What are the overall reaction orders for the reactions described in Equations 14.9 and 14.11?

(b) What are the units of the rate constant for the rate law in Equation 14.9?

order in CHCl3 and one-half order in Cl2 The overall reaction order is three halves

(b) For the rate law for Equation 14.9, we have

Units of rate = 1units of rate constant21units of concentration2so

Units of rate constant = units of rate

586 chapTer 14 chemical Kinetics

Using Initial Rates to Determine Rate Laws

We have seen that the rate law for most reactions has the general form

Rate = k3reactant 14m3reactant 24nc Thus, the task of determining the rate law becomes one of determining the reac-

tion orders, m and n In most reactions, the reaction orders are 0, 1, or 2 As noted

earlier in this section, we can use the response of the reaction rate to a change in initial concentration to determine the reaction order.

In working with rate laws, it is important to realize that the rate of a reaction depends on concentration but the rate constant does not As we will see later in this

chapter, the rate constants (and hence the reaction rate) are affected by temperature and by the presence of a catalyst.

14.10? (b) What are the units of the rate constant for Equation

14.10?

The initial rate of a reaction A + B ¡ C was measured for several different starting concentrations of

A and B, and the results are as follows:

s a m P l e

exeRCise 14.6 determining a Rate law from initial Rate data

plan (a) We assume that the rate law has the following form:

Rate = k3A4 m3B4n We will use the given data to deduce the

reac-tion orders m and n by determining how changes in the

concentra-tion change the rate (b) Once we know m and n, we can use the

rate law and one of the sets of data to determine the rate constant

k (c) Upon determining both the rate constant and the reaction

orders, we can use the rate law with the given concentrations to calculate rate

Solve

(a) If we compare experiments 1 and 2, we see that [A] is held constant

and [B] is doubled Thus, this pair of experiments shows how [B]

affects the rate, allowing us to deduce the order of the rate law with

respect to B Because the rate remains the same when [B] is doubled, the concentration of B has no effect on the reaction rate The rate law is therefore zero order in B (that is, n = 0).

In experiments 1 and 3, [B] is held constant, so these data show how

[A] affects rate Holding [B] constant while doubling [A] increases

the rate fourfold This result indicates that rate is proportional to 3A42

(that is, the reaction is second order in A) Hence, the rate law is Rate = k3A423B40 = k3A42

(b) Using the rate law and the data from experiment 1, we have k = rate

3A42 = 4.0 * 10-5 M>s

10.100 M22 = 4.0 * 10-3 M-1 s-1

(c) Using the rate law from part (a) and the rate constant from

part (b), we have Rate = k3A42= 14.0 * 10-3 M-1s-1210.050 M22= 1.0 * 10-5 M>s

Because [B] is not part of the rate law, it is irrelevant to the rate if there is at least some B present to react with A

experiment number [a] (M) [B] (M) initial Rate 1M,s2

Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the

reaction when 3A4 = 0.050M and 3B4 = 0.100 M.

solution

analyze We are given a table of data that relates concentrations of reactants with initial rates of reaction and

asked to determine (a) the rate law, (b) the rate constant, and (c) the rate of reaction for a set of

concentra-tions not listed in the table

secTIon 14.4 The change of concentration with Time 587

with time

The rate laws we have examined so far enable us to calculate the rate of a reaction from

the rate constant and reactant concentrations In this section, we will show that rate

laws can also be converted into equations that show the relationship between

concen-trations of reactants or products and time The mathematics required to accomplish

this conversion involves calculus We do not expect you to be able to perform the

calcu-lus operations, but you should be able to use the resulting equations We will apply this

conversion to three of the simplest rate laws: those that are first order overall, those that

are second order overall, and those that are zero order overall.

First-Order Reactions

A first-order reaction is one whose rate depends on the concentration of a single

re-actant raised to the first power If a reaction of the type A ¡ products is first order,

the rate law is:

Rate = - ∆3A4 ∆t = k3A4

This form of a rate law, which expresses how rate depends on concentration, is called

the differential rate law Using the operation from calculus called integration, this

rela-tionship can be transformed into an equation known as the integrated rate law for a

check A good way to check our rate law is to use the concentrations in experiment 2 or 3 and see

if we can correctly calculate the rate Using data from experiment 3, we have

Rate = k3A42= 14.0 * 10-3 M-1 s-1210.200 M22= 1.6 * 10-4 M>s

Thus, the rate law correctly reproduces the data, giving both the correct number and the correct

units for the rate

Practice exercise 1

A certain reaction X + Y ¡ Z is described as being first order in [X] and third order

overall Which of the following statements is or are true?:

(i) The rate law for the reaction is: Rate = k3X43Y42

(ii) If the concentration of X is increased by a factor of 1.5, the rate will increase by a factor

of 2.25

(iii) If the concentration of Y is increased by a factor of 1.5, the rate will increase by a factor of 2.25

(a) Only one of the statements is true

(c) Statements (i) and (iii) are true

(e) All three statements are true.

(b) Statements (i) and (ii) are true

(d) Statements (ii) and (iii) are true.

(a) Determine the rate law for this reaction (b) Calculate the rate constant.

(c) Calculate the rate when 3NO4 = 0.050 M and 3H24 = 0.150 M.

588 chapTer 14 chemical Kinetics

first-order reaction that relates the initial concentration of A, 3A40, to its concentration

at any other time t, 3A4t:

ln3A4t - ln3A40 = -kt or ln 3A40 3A4t = -kt [14.12]

The function “ln” in Equation 14.12 is the natural logarithm (Appendix A.2) Equation 14.12 can also be rearranged to

Equations 14.12 and 14.13 can be used with any concentration units as long as the units are the same for both 3A4t and 3A40.

For a first-order reaction, Equation 14.12 or 14.13 can be used in several ways

Given any three of the following quantities, we can solve for the fourth: k, t, 3A40, and

3A4t Thus, you can use these equations to determine (1) the concentration of a reactant remaining at any time after the reaction has started, (2) the time interval required for a given fraction of a sample to react, or (3) the time interval required for a reactant con- centration to fall to a certain level.

The decomposition of a certain insecticide in water at 12 °C follows

first-order kinetics with a rate constant of 1.45 yr-1 A quantity of this

insecticide is washed into a lake on June 1, leading to a

concentra-tion of 5.0 * 10-7 g>cm3 Assume that the temperature of the lake is

constant (so that there are no effects of temperature variation on the

rate) (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the insecticide concen-

tration to decrease to 3.0 * 10- 7 g>cm3?

s a m P l e

exeRCise 14.7 using the integrated First-order Rate law

solution

analyze We are given the rate constant for a reaction that obeys

first-order kinetics, as well as information about concentrations and times,

and asked to calculate how much reactant (insecticide) remains after

1 yr We must also determine the time interval needed to reach a

par-ticular insecticide concentration Because the exercise gives time in

(a) and asks for time in (b), we will find it most useful to use the

inte-grated rate law, Equation 14.13

plan

(a) We are given k = 1.45 yr-1, t = 1.00 yr, and

3insecticide40 = 5.0 * 10-7 g>cm3, and so Equation 14.13 can be solved for 3insecticide4t

(b) We have k = 1.45 yr-1, 3insecticide40 = 5.0 * 10-7 g>cm3, and 3insecticide4t = 3.0 * 10-7 g>cm3, and so we can solve Equation

14.13 for time, t.

Solve

(a) Substituting the known quantities into Equation 14.13, we have ln3insecticide4t = 1 yr = -11.45 yr-1211.00 yr2 + ln15.0 * 10-72

We use the ln function on a calculator to evaluate the second term on

the right [that is, ln15.0 * 10-724, giving ln3insecticide4t = 1 yr = -1.45 + 1-14.512 = -15.96

check In part (a) the concentration remaining after 1.00 yr (that

is, 1.2 * 10- 7 g>cm3) is less than the original concentration

15.0 * 10- 7 g>cm32, as it should be In (b) the given concentration

13.0 * 10-7 g>cm32 is greater than that remaining after 1.00 yr, ing that the time must be less than a year Thus, t = 0.35 yr is a reason-

indicat-able answer

Practice exercise 1

At 25 °C, the decomposition of dinitrogen pentoxide, N2O51g2, into NO21g2 and O21g2 follows

first-order kinetics with k = 3.4 * 10-5 s-1 A sample of N2O5 with an initial pressure of

760 torr decomposes at 25 °C until its partial pressure is 650 torr How much time (in seconds) has elapsed?

(a) 5.3 * 10-6 (b) 2000 (c) 4600 (d) 34,000 (e) 190,000

Practice exercise 2

The decomposition of dimethyl ether, 1CH322O, at 510 °C is a first-order process with a rate constant of 6.8 * 10-4 s-1:

1CH322O1g2 ¡ CH41g2 + H21g2 + CO1g2

If the initial pressure of 1CH322O is 135 torr, what is its pressure after 1420 s?

To obtain 3insecticide4t = 1 yr, we use the inverse natural logarithm, or

e x, function on the calculator: 3insecticide4t = 1 yr = e- 15.96 = 1.2 * 10- 7g>cm3

Note that the concentration units for 3A4t and 3A40 must be the same

(b) Again substituting into Equation 14.13, with

3insecticide4t = 3.0 * 10-7 g>cm3, gives ln13.0 * 10-72 = -11.45 yr-121t2 + ln15.0 * 10-72

Solving for t gives t = -3ln13.0 * 10-72 - ln15.0 * 10-724>1.45 yr-1

= -1-15.02 + 14.512>1.45 yr-1 = 0.35 yr

secTIon 14.4 The change of concentration with Time 589

Equation 14.13 can be used to verify whether a reaction is first order and to

determine its rate constant This equation has the form of the general equation for

a straight line, y = mx + b, in which m is the slope and b is the y-intercept of the

line (Appendix A.4):

For a first-order reaction, therefore, a graph of ln3A4t versus time gives a straight line

with a slope of -k and a y-intercept of ln3A40 A reaction that is not first order will not

yield a straight line.

As an example, consider the conversion of methyl isonitrile 1CH3NC2 to its isomer

acetonitrile 1CH3CN2 (▶ Figure 14.7) Because experiments show that the reaction is

first order, we can write the rate equation:

ln3CH3NC4t = -kt + ln3CH3NC40

We run the reaction at a temperature at which methyl isonitrile is a gas 1199 °C2, and

▼ Figure 14.8(a) shows how the pressure of this gas varies with time We can use

pressure as a unit of concentration for a gas because we know from the ideal-gas

law the pressure is directly proportional to the number of moles per unit volume

Figure 14.8(b) shows that a plot of the natural logarithm of the pressure versus time

is a straight line The slope of this line is -5.1 * 10-5 s-1 (You should verify this for

yourself, remembering that your result may vary slightly from ours because of

inaccu-racies associated with reading the graph.) Because the slope of the line equals -k, the

rate constant for this reaction equals 5.1 * 10-5 s-1.

Second-Order Reactions

A second-order reaction is one for which the rate depends either on a reactant

concen-tration raised to the second power or on the concenconcen-trations of two reactants each raised

to the first power For simplicity, let’s consider reactions of the type A ¡ products

or A + B ¡ products that are second order in just one reactant, A:

Rate = - ∆3A4 ∆t = k3A42

▲ Figure 14.7 The first-order reaction of

4.03.84.24.44.64.85.05.2

590 chapTer 14 chemical Kinetics

With the use of calculus, this differential rate law can be used to derive the integrated rate law for second-order reactions:

1 3A4t = kt + 1

This equation, like Equation 14.13, has four variables, k, t, 3A40, and 3A4t, and any one

of these can be calculated knowing the other three Equation 14.14 also has the form

of a straight line 1y = mx + b2 If the reaction is second order, a plot of 1>3A4t

ver-sus t yields a straight line with slope k and y-intercept 1>3A40 One way to distinguish

between first- and second-order rate laws is to graph both ln3A4t and 1>3A4t against

t If the ln3A4t plot is linear, the reaction is first order; if the 1>3A4t plot is linear, the reaction is second order.

solution

analyze We are given the concentrations of a reactant at various times

during a reaction and asked to determine whether the reaction is first

or second order

plan We can plot ln3NO24 and 1>3NO24 against time If one plot or the

other is linear, we will know the reaction is either first or second order

Solve To graph ln3NO24 and 1>3NO24 against time, we first make the

following calculations from the data given:

s a m P l e

exeRCise 14.8 determining Reaction order from the integrated Rate law

The following data were obtained for the gas-phase decomposition

of nitrogen dioxide at 300 °C, NO21g2 ¡ NO1g2 + 12 O21g2 Is the

reaction first or second order in NO2?

As ▶ Figure 14.9 shows, only the plot of 1>3NO24 versus

time is linear Thus, the reaction obeys a second-order rate law:

Rate = k3NO242 From the slope of this straight-line graph,

we determine that k = 0.543 M-1 s-1 for the disappearance

of NO2

Practice exercise 1

For a certain reaction A ¡ products, a plot of ln[A] versus

time produces a straight line with a slope of -3.0 * 10-2 s-1

Which of the following statements is or are true?:

(i) The reaction follows first-order kinetics

(ii) The rate constant for the reaction is 3.0 * 10-2 s-1

(iii) The initial concentration of [A] was 1.0 M.

secTIon 14.4 The change of concentration with Time 591

Zero-Order Reactions

We have seen that in a first-order reaction the concentration of a reactant A decreases

nonlinearly, as shown by the red curve in ▶ Figure 14.10 As [A] declines, the rate at

which it disappears declines in proportion A zero-order reaction is one in which the

rate of disappearance of A is independent of [A] The rate law for a zero-order reaction is

Rate = - ∆3A4 ∆t = k

The integrated rate law for a zero-order reaction is

3A4t = -kt + 3A40

where 3A4t is the concentration of A at time t and 3A40 is the initial concentration

This is the equation for a straight line with vertical intercept 3A40 and slope -kt, as

shown in the blue curve in Figure 14.10.

The most common type of zero-order reaction occurs when a gas undergoes

decomposition on the surface of a solid If the surface is completely covered by

decomposing molecules, the rate of reaction is constant because the number of

react-ing surface molecules is constant, so long as there is some gas-phase substance left.

Half-Life

The half-life of a reaction, t1>2, is the time required for the concentration of a

reac-tant to reach half its initial value, 3A4t1>2 = 123A40 Half-life is a convenient way to

de-scribe how fast a reaction occurs, especially if it is a first-order process A fast reaction

has a short half-life.

We can determine the half-life of a first-order reaction by substituting 3A4t1>2 = 123A40

for 3A4t and t1>2 for t in Equation 14.12:

ln

1

23A40 3A40 = -kt1>2

ln 12 = -kt1>2

t1>2 = - ln

1 2

k = 0.693 k [14.15]

From Equation 14.15, we see that t1>2 for a first-order rate law does

not depend on the initial concentration of any reactant Consequently,

the half-life remains constant throughout the reaction If, for example,

the concentration of a reactant is 0.120 M at some instant in the

reac-tion, it will be 1 210.120 M2 = 0.060 M after one half-life After one more

half-life passes, the concentration will drop to 0.030 M, and so on

Equa-tion 14.15 also indicates that, for a first-order reacEqua-tion, we can calculate

t1>2 if we know k and calculate k if we know t1>2.

The change in concentration over time for the first-order

rearrange-ment of gaseous methyl isonitrile at 199 °C is graphed in ▶ Figure 14.11

Because the concentration of this gas is directly proportional to its

pres-sure during the reaction, we have chosen to plot prespres-sure rather than

concentration in this graph The first half-life occurs at 13,600 s (3.78

h) At a time 13,600 s later, the methyl isonitrile pressure (and therefore,

concentration) has decreased to half of one-half, or one-fourth, of the initial value In a

first-order reaction, the concentration of the reactant decreases by one-half in each of a

series of regularly spaced time intervals, each interval equal to t1>2.

(a) Only one of the statements is true.

(b) Statements (i) and (ii) are true.

(c) Statements (i) and (iii) are true.

(d) Statements (ii) and (iii) are true.

(e) All three statements are true.

Practice exercise 2

The decomposition of NO2 discussed in the Sample Exercise is second order in NO2 with k = 0.543 M-1 s-1 If the initial concentration of NO2 in a closed vessel is 0.0500 M, what is the

concentration of this reactant after 0.500 h?

▲ Figure 14.11 Kinetic data for the rearrangement of methyl isonitrile to acetonitrile at 199 °c, showing the half-life of

▲ Figure 14.10 Comparison of first-order and zero-order reactions for the disappearance

of reactant A with time.

37.575150

Time (s)

592 chapTer 14 chemical Kinetics

Give It Some Thought

If a solution containing 10.0 g of a substance reacts by first-order kinetics, how many grams remain after three half-lives?

Chemistry Put to Work

Methyl Bromide in

the Atmosphere

The compounds known as chlorofluorocarbons (CFCs) are well-known

agents responsible for the destruction of Earth’s protective ozone layer

Another simple molecule that has the potential to destroy the

strato-spheric ozone layer is methyl bromide, CH3Br (▼ Figure 14.12)

Because this substance has a wide range of uses, including

antifun-gal treatment of plant seeds, it has been produced in large quantities

in the past (about 150 million pounds per year worldwide in 1997, at

the height of its production) In the stratosphere, the C ¬ Br bond is

broken through absorption of short-wavelength radiation The resultant

Br atoms then catalyze decomposition of O3.Methyl bromide is removed from the lower atmosphere by a vari-ety of mechanisms, including a slow reaction with ocean water:

CH3Br1g2 + H2O1l2 ¡ CH3OH1aq2 + HBr1aq2 [14.16]

To determine the potential importance of CH3Br in destruction

of the ozone layer, it is important to know how rapidly the reaction in Equation 14.16 and all other reactions remove CH3Br from the lower atmosphere before it can diffuse into the stratosphere

The average lifetime of CH3Br in Earth’s lower atmosphere is difficult to measure because the conditions that exist in the atmo-sphere are too complex to be simulated in the laboratory Instead, scientists analyzed nearly 4000 atmospheric samples collected above the Pacific Ocean for the presence of several trace organic substances, including methyl bromide From these measurements,

it was possible to estimate the atmospheric residence time for CH3Br.The atmospheric residence time is related to the half-life for CH3Br in the lower atmosphere, assuming CH3Br decom-poses by a first-order process From the experimental data, the half-life for methyl bromide in the lower atmosphere is estimated to be 0.8 { 0.1 yr That is, a collection of CH3Br molecules present at any given time will, on average, be 50% decomposed after 0.8 yr, 75% decomposed after 1.6 yr, and so

on A half-life of 0.8 yr, while comparatively short, is still ficiently long so that CH3Br contributes significantly to the de-struction of the ozone layer

suf-In 1997 an international agreement was reached to phase out use of methyl bromide in developed countries by 2005 How-ever, in recent years exemptions for critical agricultural use have been requested and granted Nevertheless, authorized worldwide production was down to 26 million pounds in 2012, three-fourths

of which is used in the United States

analyze We are asked to estimate the half-life of a reaction from a

graph of concentration versus time and then to use the half-life to

cal-culate the rate constant for the reaction

plan

(a) To estimate a half-life, we can select a concentration and then

determine the time required for the concentration to decrease to

half of that value

s a m P l e

exeRCise 14.9 determining the half-life of a First-order Reaction

The reaction of C4H9Cl with water is a first-order reaction (a) Use Figure 14.4 to estimate the

half-life for this reaction (b) Use the half-half-life from (a) to calculate the rate constant.

(b) Equation 14.15 is used to calculate the rate constant from the

half-life

Solve

(a) From the graph, we see that the initial value of 3C4H9Cl4 is 0.100 M

The half-life for this first-order reaction is the time required for 3C4H9Cl4 to decrease to 0.050 M, which we can read off the

graph This point occurs at approximately 340 s

secTIon 14.5 Temperature and rate 593

The half-life for second-order and other reactions depends on reactant

concentra-tions and therefore changes as the reaction progresses We obtained Equation 14.15 for

the half-life for a first-order reaction by substituting 3A4t1>2 = 123A40 for 3A4t and t1>2

for t in Equation 14.12 We find the half-life of a second-order reaction by making the

same substitutions into Equation 14.14:

11

23A40 = kt1>2 +

1 3A40 2

3A40

-1 3A40 = kt1>2

t1>2 = 1

In this case, the half-life depends on the initial concentration of reactant—the lower the

initial concentration, the longer the half-life.

Give It Some Thought

Why can we report the half-life for a first-order reaction without knowing the

initial concentration, but not for a second-order reaction?

The rates of most chemical reactions increase as the temperature rises For example,

dough rises faster at room temperature than when refrigerated, and plants grow more

rapidly in warm weather than in cold We can see the effect of temperature on reaction

rate by observing a chemiluminescence reaction (one that produces light), such as that

in Cyalume® light sticks (▶ Figure 14.13).

How is this experimentally observed temperature effect reflected in the rate law?

The faster rate at higher temperature is due to an increase in the rate constant with

increasing temperature For example, let’s reconsider the first-order reaction we saw in

Figure 14.7, namely CH3NC ¡ CH3CN Figure 14.14 shows the rate constant for

this reaction as a function of temperature The rate constant and, hence, the rate of the

reaction increase rapidly with temperature, approximately doubling for each 10 °C rise.

The Collision Model

Reaction rates are affected both by reactant concentrations and by temperature The

collision model, based on the kinetic-molecular theory (Section 10.7), accounts for

both of these effects at the molecular level The central idea of the collision model is

that molecules must collide to react The greater the number of collisions per second,

the greater the reaction rate As reactant concentration increases, therefore, the number

of collisions increases, leading to an increase in reaction rate According to the

kinetic-molecular theory of gases, increasing the temperature increases kinetic-molecular speeds As

molecules move faster, they collide more forcefully (with more energy) and more

fre-quently, increasing reaction rates.

(b) Solving Equation 14.15 for k, we have

k = 0.693t

1>2 = 0.693

340 s = 2.0 * 10-3 s-1

check At the end of the second half-life, which should occur at 680 s,

the concentration should have decreased by yet another factor of 2, to

0.025 M Inspection of the graph shows that this is indeed the case.

Practice exercise 1

We noted in an earlier Practice Exercise that at 25 °C the

decom-position of N2O51g2 into NO21g2 and O21g2 follows first-order

kinetics with k = 3.4 * 10-5 s-1 How long will it take for a sample originally containing 2.0 atm of N2O5 to reach a partial pressure of 380 torr?

(a) 5.7 h (b) 8.2 h (c) 11 h (d) 16 h (e) 32 h

Practice exercise 2

(a) Using Equation 14.15, calculate t1>2 for the decomposition of the insecticide described in Sample Exercise 14.7

(b) How long does it take for the concentration of the insecticide

to reach one-quarter of the initial value?

Hot water Cold water

▲ Figure 14.13 Temperature affects the rate of the chemiluminescence reaction in light sticks: The chemiluminescent reaction occurs more rapidly in hot water, and more light is produced.

Go FiGuRe

Why does the light stick glow with less light in cold water than in hot water?

594 chapTer 14 chemical Kinetics

For a reaction to occur, though, more is required than simply a collision—it must be the right kind of collision For most reactions, in fact, only a tiny fraction of collisions leads to a reaction For example, in a mixture of H2 and I2 at ordinary tem- peratures and pressures, each molecule undergoes about 1010 collisions per second If every collision between H2 and I2 resulted in the formation of HI, the reaction would

be over in much less than a second Instead, at room temperature the reaction proceeds very slowly because only about one in every 1013 col- lisions produces a reaction What keeps the reaction from occurring more rapidly?

The Orientation Factor

In most reactions, collisions between molecules result in a chemical tion only if the molecules are oriented in a certain way when they collide The relative orientations of the molecules during collision determine whether the atoms are suitably positioned to form new bonds For ex- ample, consider the reaction

reac-Cl + NOreac-Cl ¡ NO + reac-Cl2 which takes place if the collision brings Cl atoms together to form Cl2,

as shown in the top panel of ▼ Figure 14.15 In contrast, in the collision shown in the lower panel, the two Cl atoms are not colliding directly with one another, and no products are formed.

Activation Energy Molecular orientation is not the only factor influencing whether a molecular collision will produce a reaction In 1888 the Swedish chemist Svante Arrhenius suggested that molecules must possess a certain minimum amount of energy to react According to the collision model, this energy comes from the kinetic energies of the colliding mol- ecules Upon collision, the kinetic energy of the molecules can be used to stretch, bend, and ultimately break bonds, leading to chemical reactions That is, the kinetic energy

is used to change the potential energy of the molecule If molecules are moving too slowly—in other words, with too little kinetic energy—they merely bounce off one an- other without changing The minimum energy required to initiate a chemical reaction

is called the activation energy, Ea, and its value varies from reaction to reaction.

▲ Figure 14.14 Temperature dependence

of the rate constant for methyl isonitrile

conversion to acetonitrile The four points

indicated are used in Sample Exercise 14.11

Go FiGuRe

Would you expect this curve to

eventually go back down to lower

values? Why or why not?

▲ Figure 14.15 Molecular collisions may or may not lead to a chemical reaction between Cl and NOCl.

Effective collision, reaction occurs,

Cl2 forms

Ineffective collision,

no reaction possible,

no Cl2

secTIon 14.5 Temperature and rate 595

The situation during reactions is analogous to that shown in ▲ Figure 14.16

The golfer hits the ball to make it move over the hill in the direction of the cup

The hill is a barrier between ball and cup To reach the cup, the player must impart

enough kinetic energy with the putter to move the ball to the top of the barrier If

he does not impart enough energy, the ball will roll partway up the hill and then

back down toward him In the same way, molecules require a certain minimum

energy to break existing bonds during a chemical reaction We can think of this

minimum energy as an energy barrier In the rearrangement of methyl isonitrile

to acetonitrile, for example, we might imagine the reaction passing through an

intermediate state in which the N ‚ C portion of the methyl isonitrile molecule is

sideways:

N

▶ Figure 14.17 shows that energy must be supplied to stretch the bond between the

H3C group and the N ‚ C group to allow the N ‚ C group to rotate After the N ‚ C

group has twisted sufficiently, the C ¬ C bond begins to form, and the energy of the

molecule drops Thus, the barrier to formation of acetonitrile represents the energy

necessary to force the molecule through the relatively unstable

inter-mediate state, analogous to forcing the ball in Figure 14.16 over

the hill The difference between the energy of the starting molecule

and the highest energy along the reaction pathway is the activation

energy, Ea The molecule having the arrangement of atoms shown

at the top of the barrier is called either the activated complex or the

transition state.

The conversion of H3C ¬ N ‚ C to H3C ¬ C ‚ N is

exother-mic Figure 14.17 therefore shows the product as having a lower

energy than the reactant The energy change for the reaction, ∆E,

has no effect on reaction rate, however The rate depends on the

magnitude of Ea; generally, the lower the value of Ea is, the faster the

reaction.

Notice that the reverse reaction is endothermic The activation

energy for the reverse reaction is equal to the energy that must be

overcome if approaching the barrier from the right: ∆E + Ea Thus,

to reach the activated complex for the reverse reaction requires more energy than for

the forward reaction—for this reaction, there is a larger barrier to overcome going from

right to left than from left to right.

▲ Figure 14.16 Energy is needed to overcome a barrier between initial and final states.

N

H3C CC

Activated complex forms

C—C bond forms

▲ Figure 14.17 Energy profile for conversion of methyl isonitrile 1h 3 cnc2 to its isomer acetonitrile 1h 3 ccn2.

Go FiGuRe

How does the energy needed to overcome the energy barrier compare with the overall change in energy for this reaction?

596 chapTer 14 chemical Kinetics

Give It Some Thought

Suppose you could measure the rates for both the forward and reverse reactions of the process in Figure 14.17 In which direction would the rate

be larger? Why?

Any particular methyl isonitrile molecule acquires sufficient energy to overcome the energy barrier through collisions with other molecules Recall from the kinetic- molecular theory of gases that, at any instant, gas molecules are distributed in energy over a wide range (Section 10.7)▼ Figure 14.18 shows the distribution of kinetic energies for two temperatures, comparing them with the minimum energy needed for

reaction, Ea At the higher temperature a much greater fraction of the molecules have

kinetic energy greater than Ea, which leads to a greater rate of reaction.

Give It Some Thought

Suppose we have two reactions, A ¡ B and B ¡ C You can isolate B, and

it is stable Is B the transition state for the reaction A ¡ C?

For a collection of molecules in the gas phase, the fraction of molecules that have kinetic

energy equal to or greater than Ea is given by the expression

In this equation, R is the gas constant 18.314 J>mol@K2 and T is the absolute temperature To get an idea of the magnitude of f, let’s sup- pose that Ea is 100 kJ>mol, a value typical of many reactions, and that

T is 300 K The calculated value of f is 3.9 * 10-18, an extremely small

number! At 320 K, f = 4.7 * 10-17 Thus, only a 20° increase in perature produces a more than tenfold increase in the fraction of mol- ecules possessing at least 100 kJ>mol of energy.

tem-The Arrhenius Equation Arrhenius noted that for most reactions the increase in rate with increasing temperature is nonlinear (Figure 14.14) He found that most reaction-rate data obeyed an equation based on (a) the frac-

tion of molecules possessing energy Ea or greater, (b) the number of collisions per second, and (c) the fraction of collisions that have the appropriate orientation

These three factors are incorporated into the Arrhenius equation:

In this equation, k is the rate constant, Ea is the activation energy, R is the gas

con-stant 18.314 J>mol@K2, and T is the absolute temperature The frequency factor, A, is

constant, or nearly so, as temperature is varied This factor is related to the frequency

of collisions and the probability that the collisions are favorably oriented for

reac-tion.* As the magnitude of Ea increases, k decreases because the fraction of molecules that possess the required energy is smaller Thus, at fixed values of T and A, reaction rates decrease as Ea increases.

▲ Figure 14.18 The effect of temperature

on the distribution of kinetic energies of

molecules in a sample.

Go FiGuRe

What would the curve look like for a

temperature higher than that for the

red curve in the figure?

secTIon 14.5 Temperature and rate 597

Determining the Activation Energy

We can calculate the activation energy for a reaction by manipulating the Arrhenius

equation Taking the natural log of both sides of Equation 14.19, we obtain

which has the form of the equation for a straight line A graph of ln k versus 1>T is a

line with a slope equal to -Ea>R and a y-intercept equal to ln A Thus, the activation

energy can be determined by measuring k at a series of temperatures, graphing ln k

versus 1>T, and calculating Ea from the slope of the resultant line.

We can also use Equation 14.20 to evaluate Ea in a nongraphical way if we know

the rate constant of a reaction at two or more temperatures For example, suppose that

at two different temperatures T1 and T 2 a reaction has rate constants k1 and k2 For each

condition, we have

ln k1 = - RT Ea

1 + ln A and ln k2 = - Ea

RT2 + ln A Subtracting ln k2 from ln k1 gives

ln k1 - ln k2 = a- RT Ea

1 + ln Ab - a- RT Ea

2 + ln Ab

solution

The lower the activation energy, the faster the reaction The value of

∆E does not affect the rate Hence, the order from slowest reaction to

fastest is 2 6 3 6 1

Practice exercise 1

Which of the following statements is or are true?

(i) The activation energies for the forward and reverse directions

of a reaction can be different

(ii) Assuming that A is constant, if both E a and T increase, then k

will increase

s a m P l e

exeRCise 14.10 activation energies and speeds of Reaction

Consider a series of reactions having these energy profiles:

(iii) For two different reactions, the one with the smaller value of

E a will necessarily have the larger value for k

(a) Only one of the statements is true

(b) Statements (i) and (ii) are true

(c) Statements (i) and (iii) are true

(d) Statements (ii) and (iii) are true

(e) All three statements are true.

598 chapTer 14 chemical Kinetics

Simplifying this equation and rearranging gives

ln k1 k

2 = Ea

Equation 14.21 provides a convenient way to calculate a rate constant k1 at some

tem-perature T1 when we know the activation energy and the rate constant k2 at some other

temperature T2.

Solve

(a) We must first convert the temperatures from degrees

Celsius to kelvins We then take the inverse of each

tem-perature, 1>T, and the natural log of each rate constant,

ln k This gives us the table shown at the right:

A graph of ln k versus 1>T is a straight line (▼ Figure 14.19).

The slope of the line is obtained by choosing any two

well-separated points and using the coordinates of each:

Slope = ∆∆x y = -6.6 - 1-10.42

0.00195 - 0.00215 = -1.9 * 104

Because logarithms have no units, the numerator in this

equa-tion is dimensionless The denominator has the units of 1>T,

namely, K- 1 Thus, the overall units for the slope are K The slope

equals -E a >R We use the value for the gas constant R in units of

J>mol@K (Table 10.2) We thus obtain

Slope = - E R a

E a = -1slope21R2 = -1-1.9 * 104 K2a8.314mol@K b aJ 1000 J b1 kJ

= 1.6 * 102 kJ>mol = 160 kJ>mol

We report the activation energy to only two significant figures

because we are limited by the precision with which we can

read the graph in Figure 14.19

s a m P l e

exeRCise 14.11 determining the activation energy

The following table shows the rate constants for the rearrangement of methyl isonitrile at various

temperatures (these are the data points in Figure 14.14):

analyze We are given rate constants, k, measured at several temperatures and asked to determine the

activation energy, E a , and the rate constant, k, at a particular temperature.

plan We can obtain E a from the slope of a graph of ln k versus 1>T Once we know E a, we can use

Equation 14.21 together with the given rate data to calculate the rate constant at 430.0 K

T 1K2 1,T 1K −12 ln k

462.9 2.160 * 10-3 -10.589472.1 2.118 * 10-3 -9.855503.5 1.986 * 10-3 -7.370524.4 1.907 * 10-3 -5.757

▲ Figure 14.19 Graphical determination of activation energy E a.

secTIon 14.6 reaction Mechanisms 599

A balanced equation for a chemical reaction indicates the substances present at the start

of the reaction and those present at the end of the reaction It provides no

informa-tion, however, about the detailed steps that occur at the molecular level as the reactants

are turned into products The steps by which a reaction occurs is called the reaction

mechanism At the most sophisticated level, a reaction mechanism describes the order

in which bonds are broken and formed and the changes in relative positions of the

at-oms in the course of the reaction.

Elementary Reactions

We have seen that reactions take place because of collisions between reacting

mole-cules For example, the collisions between molecules of methyl isonitrile 1CH3NC2 can

provide the energy to allow the CH3NC to rearrange to acetonitrile:

N

Similarly, the reaction of NO and O3 to form NO2 and O2 appears to occur as a result

of a single collision involving suitably oriented and sufficiently energetic NO and O3

molecules:

NO1g2 + O31g2 ¡ NO21g2 + O21g2 [14.22]

Both reactions occur in a single event or step and are called elementary reactions.

The number of molecules that participate as reactants in an elementary reaction

defines the molecularity of the reaction If a single molecule is involved, the reaction is

unimolecular The rearrangement of methyl isonitrile is a unimolecular process

Ele-mentary reactions involving the collision of two reactant molecules are bimolecular

The reaction between NO and O3 is bimolecular Elementary reactions involving the

simultaneous collision of three molecules are termolecular Termolecular reactions are

far less probable than unimolecular or bimolecular processes and are extremely rare

The chance that four or more molecules will collide simultaneously with any regularity

is even more remote; consequently, such collisions are never proposed as part of a

reac-tion mechanism Thus, nearly all reacreac-tion mechanisms contain only unimolecular and

bimolecular elementary reactions.

(b) To determine the rate constant, k1, at T1= 430.0 K, we can

use Equation 14.21 with E a = 160 kJ>mol and one of the

rate constants and temperatures from the given data, such as

To one significant figure, what is the value for the frequency factor

A for the data presented in Sample Exercise 14.11.

600 chapTer 14 chemical Kinetics

Give It Some Thought

What is the molecularity of the elementary reaction?

NO21g2 + CO1g2 ¡ NO1g2 + CO21g2 [14.23]

appears to proceed in two elementary reactions (or two elementary steps), each of which

is bimolecular First, two NO2 molecules collide, and an oxygen atom is transferred from one to the other The resultant NO3 then collides with a CO molecule and trans- fers an oxygen atom to it:

NO21g2 + NO21g2 ¡ NO31g2 + NO1g2 NO31g2 + CO1g2 ¡ NO21g2 + CO21g2 Thus, we say that the reaction occurs by a two-step mechanism.

The chemical equations for the elementary reactions in a multistep anism must always add to give the chemical equation of the overall process In

mech-the present example, mech-the sum of mech-the two elementary reactions is

2 NO21g2 + NO31g2 + CO1g2 ¡ NO21g2 + NO31g2 + NO1g2 + CO21g2 Simplifying this equation by eliminating substances that appear on both sides gives Equation 14.23, the net equation for the process.

Because NO3 is neither a reactant nor a product of the reaction—it is formed in one elementary reaction and consumed in the next—it is called

an intermediate Multistep mechanisms involve one or more intermediates

Intermediates are not the same as transition states, as shown in ◀ Figure 14.20 Intermediates can be stable and can therefore sometimes be identified and even isolated Transition states, on the other hand, are always inherently unstable and as such can never be isolated Nevertheless, the use of advanced “ultrafast” techniques sometimes allows us to characterize them.

Reaction progressReactants

Intermediate

Products

Transitionstate

Transition

state

▲ Figure 14.20 The energy profile of a

reaction, showing transition states and an

intermediate.

Go FiGuRe

For this profile, is it easier for a

molecule of the intermediate to

convert to reactants or products?

solution

analyze We are given a two-step mechanism and asked for (a) the

molecularities of each of the two elementary reactions, (b) the

equa-tion for the overall process, and (c) the intermediate.

plan The molecularity of each elementary reaction depends on the

number of reactant molecules in the equation for that reaction The

overall equation is the sum of the equations for the elementary

re-actions The intermediate is a substance formed in one step of the

s a m P l e

exeRCise 14.12 determining molecularity and identifying intermediates

It has been proposed that the conversion of ozone into O2 proceeds by a two-step mechanism:

O31g2 ¡ O21g2 + O1g2

O31g2 + O1g2 ¡ 2 O21g2

(a) Describe the molecularity of each elementary reaction in this mechanism.

(b) Write the equation for the overall reaction

(c) Identify the intermediate(s).

mechanism and used in another and therefore not part of the tion for the overall reaction

equa-Solve

(a) The first elementary reaction involves a single reactant and is

consequently unimolecular The second reaction, which involves two reactant molecules, is bimolecular

secTIon 14.6 reaction Mechanisms 601

Rate Laws for Elementary Reactions

In Section 14.3, we stressed that rate laws must be determined experimentally; they cannot

be predicted from the coefficients of balanced chemical equations We are now in a position

to understand why this is so Every reaction is made up of a series of one or more

elemen-tary steps, and the rate laws and relative speeds of these steps dictate the overall rate law for

the reaction Indeed, the rate law for a reaction can be determined from its mechanism, as

we will see shortly, and compared with the experimental rate law Thus, our next challenge

in kinetics is to arrive at reaction mechanisms that lead to rate laws consistent with those

observed experimentally We start by examining the rate laws of elementary reactions.

Elementary reactions are significant in a very important way: If a reaction is

ele-mentary, its rate law is based directly on its molecularity For example, consider the

uni-molecular reaction

A ¡ products

As the number of A molecules increases, the number that reacts in a given time interval

increases proportionally Thus, the rate of a unimolecular process is first order:

Rate = k3A4

For bimolecular elementary steps, the rate law is second order, as in the reaction

A + B ¡ products Rate = k3A43B4

The second-order rate law follows directly from collision theory If we double the

con-centration of A, the number of collisions between the molecules of A and B doubles;

likewise, if we double [B], the number of collisions between A and B doubles

There-fore, the rate law is first order in both [A] and [B] and second order overall.

The rate laws for all feasible elementary reactions are given in ▼ table 14.3 Notice

how each rate law follows directly from the molecularity of the reaction It is important

(b) Adding the two elementary reactions gives

2 O31g2 + O1g2 ¡ 3 O21g2 + O1g2

Because O(g) appears in equal amounts on both sides of the equation,

it can be eliminated to give the net equation for the chemical process:

2 O31g2 ¡ 3 O21g2

(c) The intermediate is O(g) It is neither an original reactant nor a

final product but is formed in the first step of the mechanism and

consumed in the second

Practice exercise 1

Consider the following two-step reaction mechanism:

A1g2 + B1g2 ¡ X1g2 + Y1g2

X1g2 + C1g2 ¡ Y1g2 + Z1g2

Which of the following statements about this mechanism is or are true?

(i) Both of the steps in this mechanism are bimolecular.

(ii) The overall reaction is A1g2 + B1g2 + C1g2 ¡ Y1g2 + Z1g2.

(iii) The substance X(g) is an intermediate in this mechanism.

(a) Only one of the statements is true

(b) Statements (i) and (ii) are true

(c) Statements (i) and (iii) are true

(d) Statements (ii) and (iii) are true

(e) All three statements are true.

Practice exercise 2

For the reactionMo1CO26 + P1CH323¡ Mo1CO25P1CH323+ COthe proposed mechanism is

Mo1CO26¡ Mo1CO25 + COMo1CO25 + P1CH323 ¡ Mo1CO25P1CH323

(a) Is the proposed mechanism consistent with the equation for

the overall reaction? (b) What is the molecularity of each step of the mechanism? (c) Identify the intermediate(s).

table 14.3 elementary Reactions and their Rate laws